Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.12 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.12

Integrate the following with respect to x.

Question 1.
(i) \(\sqrt{x^{2}+2 x+10}\)
Answer:

(ii) \(\sqrt{x^{2}-2 x-3}\)
Answer:

(iii) \(\sqrt{(6-x)(x-4)}\)
Answer:

Question 2.
(i) \(\sqrt{9-(2 x+5)^{2}}\)
Answer:

(ii) \(\sqrt{81+(2 x+1)^{2}}\)
Answer:

(iii) \(\sqrt{(x+1)^{2}-4}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6

Question 1.
\(\frac{x}{\sqrt{1+x^{2}}}\)
Answer:
Put 1 + x² = u
2x dx = du
x dx = \(\frac{1}{2}\) du

Question 2.
\(\frac{x^{2}}{1+x^{6}}\)
Answer:

Put x³ = u
3x² dx = du
x² dx = \(\frac{1}{3}\) du

Question 3.
\(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)
Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.
\(\frac{\cot x}{\log (\sin x)}\)
Answer:

Question 7.

Answer:

Question 8.

Answer:

Put a² + b² sin²x = u
(0 + b² × 2 sin x cos x) dx = du
b² sin 2x dx = du
sin 2x dx = \(\frac{1}{b^{2}}\) du

Question 9.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:

Question 10.
\(\frac{\sqrt{x}}{1+\sqrt{x}}\)
Answer:

Question 11.

Answer:

Question 12.
α β x^α-1 e^{-β xα}
Answer:
∫α β x^α-1 e^{-β xα}
Put β x^α = u
α β x^α-1 dx = du

Question 13.
\(\tan x \sqrt{\sec x}\)
Answer:

put cos x = u
– sin x dx = du
sin x dx = – du

Question 14.
x (1 – x)¹⁷
Answer:
∫x (1 – x)¹⁷ dx
put 1 – x = u
-dx = du

Question 15.
sin⁵ x cos³ x
Answer:
∫ sin⁵ x cos³ x dx = ∫ sin⁵ x cos² x . cos x dx
= ∫ sin⁵ x (1 – sin² x) . cos x dx
= ∫ (sin⁵ x – sin⁷ x) . cos x dx
= ∫ sin⁵ x cos x dx – ∫ sin⁷ x cos x dx
put u = sin x
du = cos x dx

Question 16.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.11

Integrate the following with respect to x:

Question 1.
(i) \(\frac{2 x-3}{x^{2}+4 x-12}\)
Answer:
Let 2x – 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 4x – 12) + B
2x – 3 = A (2x + 4) + B
2x – 3 = 2Ax + 4A + B
2A = 2 ⇒ A = 1
4A + B = – 3
4 × 1 + B – 3 ⇒ B = – 3 – 4 = – 7
2x – 3 = 1 (2x + 4) – 7

(ii) \(\frac{5 x-2}{2+2 x+x^{2}}\)
Answer:
Let 5x – 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 2x + 2) + B
5x – 2 = A (2x + 2) + B
5x – 2 = 2Ax + 2A + B
2A = 5 ⇒ A = \(\frac{5}{2}\)
2A + B = -2
2 × \(\frac{5}{2}\) + B = -2 ⇒ B = -2 – 5 = -7
5x – 2 = \(\frac{5}{2}\) (2x + 2) – 7

Put x² + 2x + 12 = t
(2x + 2) dx = dt

Put x + 1 = u
dx = du

(iii) \(\frac{3 x+1}{2 x^{2}-2 x+3}\)
Answer:
Let 3x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (2x² – 2x + 3) + B
3x + 1 = A (4x – 2) + B
3x + 1 = 4Ax – 2A + B

Question 2.
(i) \(\frac{2 x+1}{\sqrt{9+4 x-x^{2}}}\)
Answer:
Let 2x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (9 + 4x – x²) + B
2x + 1 = A(4 – 2x) + B
2x + 1 = 4A – 2Ax + B
-2A = 2 ⇒ A = -1
4A + B = 1 ⇒ 4 (-1) + B = 1
B = 1 + 4 = 5
B = 5
2x + 1 = – 1 (4 – 2x) + 5

Put 9 + 4x – x² = t
(4x – 2) dx = dt

Put x – 2 = u
dx = du

(ii) \(\frac{x+2}{\sqrt{x^{2}-1}}\)
Answer:
Let x + 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² – 1) + B
x + 2 = A (2x) + B
2A = 1 ⇒ A = \(\frac{1}{2}\)
B = 2

(iii) \(\frac{2 x+3}{\sqrt{x^{2}+4 x+1}}\)
Answer:
Let 2x + 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 4x + 1) + B
2x + 3 = A (2x + 3) + B
2A = 2 ⇒ A = 1
4A + B = 3
4 × 1 + B = 3 ⇒ B = 3 – 4
B = -1
(2x + 3) = (2x + 4) – 1

Put x² + 4x + 1 = t
(2x + 4) dx = dt

Put x + 2 = u
dx = dt

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.10

Find the integrals of the following:

Question 1.
(i) \(\frac{1}{4-x^{2}}\)
Answer:

(ii) \(\frac{1}{25-4 x^{2}}\)
Answer:

(iii) \(\frac{1}{9 x^{2}-4}\)
Answer:

Question 2.
(i) \(\frac{1}{6 x-7-x^{2}}\)
Answer:

(ii) \(\frac{1}{(x+1)^{2}-25}\)
Answer:

(iii) \(\frac{1}{\sqrt{x^{2}+4 x+2}}\)
Answer:

Question 3.
(i) \(\frac{1}{\sqrt{(2+x)^{2}-1}}\)
Answer:

(ii) \(\frac{1}{\sqrt{x^{2}-4 x+5}}\)
Answer:

(iii) \(\frac{1}{\sqrt{9+8 x-x^{2}}}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.9

Integrate the following with respect to x:

Question 1.
e^x (tan x + log sec x)
Answer:
Let I = ∫e^x (tan x + log sec x) dx
Take f(x) = log sec x
f'(x) = \(\frac{1}{\sec x}\) sec x tan x
f'(x) = tan x
[∫e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^x log |sec x| + c

Question 2.

Answer:

[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]

Question 3.
e^x sec x (1 + tan x)
Answer:
Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)
Take f(x) = sec x
f ‘ (x) = sec x tan x
This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e^x f(x) + c
= e^x sec x + c

Question 4.

Answer:

Take f(x) = tan x
f'(x) = sec² x
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
I = e^x tan x + c

Question 5.

Answer:

Put tan^-1 x = t
\(\frac{1}{1+x^{2}}\)dx = dt
x = tan t
∴ I = ∫e^t [1 + tan t + tan²t] dt
= ∫e^t [1 + tan² t + tan t] dt
= ∫e^t [sec²t + tan t] dt
= ∫e^t [tan t + sec²t] dt
f(x) = tan t
f'(x) = sec²t
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^t tan t + c
I = e^tan-1(x) . x + c
I = x e^tan-1(x) + c

Question 6.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.8

Integrate the following with respect to x:

Question 1.
(i) e^ax cos bx
Answer:

(ii) e^2x sin x
Answer:

(iii) e^-x cos 2x
Answer:

Question 2.
(i) e^-3x sin 2x
Answer:

(ii) e^-4x sin 2x
Answer:

(iii) e^-3x cos x
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.7 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.7

Integrate the following with respect to x:

Question 1.
(i) 9x e^3x
Answer:

(ii) x sin 3x
Answer:

(iii) 25 x e^-5x
Answer:

(iv) x sec x tan x
Answer:
∫ x sec x tan x dx
u = x, u’ = 1 , u” = 0
dv = sec x tan x dx , v = ∫ sec x tan x dx = sec x
v₁ = ∫ v dx = ∫ sec x dx = log |sec x + tan x|
v₂ = ∫ v₁ dx = ∫ log |sec x + tan x| dx
∫ u dv = uv – u’ v₁ + u” v₂ – u”’ v₃ + ………….
∫x sec x tan x dx = x sec x – 1 × log |sec x + tan x| + 0 × ∫ log |x sec x tan x| + c
∫x sec x tan x dx = x sec x – log |sec x + tan x| + c

Question 2.
(i) x log x
Answer:

(ii) 27 x² e^3x
Answer:

(iii) x² cos x
Answer:

(iv) x³ sin x
Answer:
∫ x³ sin x
u = x³, u’ = 3x², u” = 6x , u”’ = 6 , u’^v = 0
dv = sin x dx ⇒ v = ∫ sin x dx = – cosx
v₁ = ∫v dx = ∫- cos x dx = – ∫ cos x dx = – sin x
v₂ = ∫ v₁ dx = ∫ – sin x dx = – ∫ sin x dx = – (- cos x) = cos x
v₃ = ∫ v₂ dx = ∫ cos x dx = sin x
v₄ = ∫ v₃ dx = ∫ sin x dx = – cos x
∫ u dv = uv – u’ v₁ + u” v₂ – u”’ v₃ + u’^vv₄ – ………..
∫x³ sin x dx = x³ (- cos x) – 3x² (- sin x) + 6x (cos x) – 6 sin x + 0 (- cos x) + c
= -x³ cos x + 3x² sin x + 6x cos x – 6 sin x + c

Question 3.
(i)
Answer:

(ii) x⁵ e^x2
Answer:

(iii)
Answer:

(iv)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.

(1) \(\frac{\pi}{180}\) cos x°
(2) \(\frac{1}{90}\) cos x°
(3) \(\frac{\pi}{90}\) cos x°
(4) \(\frac{2}{\pi}\) cos x°
Answer:
(2) \(\frac{1}{90}\) cos x°

Explaination:

Question 2.
If y = f(x² + 2) and f'(3) = 5 , then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at x = 1 is
(1) 5
(2) 25
(3) 15
(4) 10
Answer:
(4) 10

Explaination:
y = f(x² + 2)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f’ (x² + 2) × 2x
\(\frac{\mathrm{dy}}{\mathrm{d} x} / x=1\) = f’ (1² + 2) × 2 × 1
= f’(3) × 2
= 5 × 2 = 10

Question 3.
If y = \(\frac{1}{4}\)u⁴, u = \(\frac{2}{3}\) x³ + 5, then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) is
(1) \(\frac{1}{27}\) x² (2x³ + 15)³
(2) \(\frac{2}{27}=\) x (2x³ + 5)³
(3) \(\frac{2}{27}=\) x² (2x³ + 15)³
(4) – \(\frac{2}{27}=\) x (2x³ + 5)³
Answer:
(3) \(\frac{2}{27}=\) x² (2x³ + 15)³

Explaination:

Question 4.
If f(x) = x² – 3x, then the points at which f (x) = f’ (x) are
(1) both positive integers
(2) both negative integers
(3) both irrational
(4) one rational and another irrational
Answer:
(3) both irrational

Explaination:
f(x) = x² – 3x
f’(x) = 2x – 3
f(x) = f'(x)
⇒ x² – 3x = 2x – 3
x² – 3x – 2x + 3 = 0
x² – 5x + 3 = 0

Question 5.
If y = \(\frac{1}{a-z}\), then \(\frac{\mathrm{d} \mathrm{z}}{\mathrm{d} \mathrm{y}}\) is
(1) (a – z)²
(2) – (z – a)²
(3) (z + a)²
(4) – (z + a)²
Answer:
(1) (a – z)²

Explaination:

Question 6.
If y = cos (sin x²), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at x = \(\sqrt{\frac{\pi}{2}}\) is
(1) – 2
(2) 2
(3) – 2 \(\sqrt{\frac{\pi}{2}}\)
(4) 0
Answer:
(4) 0

Explaination:

Question 7.
If y = mx + c and f(0) = f’(0) = 1, then f(2) is
(1) 1
(2) 2
(3) 3
(4) – 3
Answer:
(3) 3

Explaination:
y = mx+c
\(\frac{d y}{d x}\) = m
y = x + c (i.e.) f(x) = x + c
y(a tx = 0) = f(0) 0 + c = 1 ⇒ c = 1
y = x + 1 ⇒ f(x) = x + 1
f(2) = 2 + 1 = 3

Question 8.
If f(x) = x tan^-1 x then f'(x) is
(1) \(1+\frac{\pi}{4}\)
(2) \(\frac{1}{2}+\frac{\pi}{4}\)
(3) \(\frac{1}{2}-\frac{\pi}{4}\)
(4) 2
Answer:
(2) \(\frac{1}{2}+\frac{\pi}{4}\)

Explaination:

Question 9.
\(\frac{d}{d x}\) (e^{x + 5 log x}) is
(1) e^x . x⁴ (x + 5)
(2) e^x . x (x + 5)
(3) e^x + \(\frac{5}{x}\)
(4) e^x – \(\frac{5}{x}\)
Answer:
(1) e^x . x⁴ (x + 5)

Explaination:

Question 10.
If the derivative of (ax – 5)e at x = 0 is – 13 , then the value of a ¡s
(1) 8
(2) – 2
(3) 5
(4) 2
Answer:
(4) 2

Explaination:
y = (ax – 5)e^3x
\(\frac{d y}{d x}\) = y’ = (ax – 5) (3e^3x) + e^3x (a)
= e^3x[3ax – 15 + a]
Given \(\frac{d y}{d x}\) = -13 at x = 0
⇒ [-15 + a] = -13
⇒ a = -13 + 15
a = 2

Question 11.
x = \(\frac{1-t^{2}}{1+t^{2}}\), y = \(\frac{2 t}{1+t^{2}}\) then \(\frac{d y}{d x}\) is
(1) – \(\frac{\mathbf{y}}{x}\)
(2) \(\frac{\mathbf{y}}{x}\)
(3) – \(\frac{x}{y}\)
(4) \(\frac{x}{y}\)
Answer:
(3) – \(\frac{x}{y}\)

Explaination:

Question 12.
If x = a sin θ and y = b cos θ, then \(\) is
(1) \(\frac{\mathbf{a}}{\mathbf{b}^{2}}\) sec² θ
(2) \(-\frac{\mathbf{b}}{\mathbf{a}}\) sec² θ
(3) \(-\frac{b}{a^{2}}\) sec³ θ
(4) \(-\frac{b^{2}}{a^{2}}\) sec³ θ
Answer:
(3) \(-\frac{b}{a^{2}}\) sec³ θ

Explaination:

Question 13.
The differential coefficient of log₁₀ x with respect to log₁₀ x is
(1) 1
(2) – (log₁₀ x)²
(3) (log₁₀ x)²
(4) \(\frac{x^{2}}{100}\)
Answer:
(2) – (log₁₀ x)²

Explaination:

Question 14.
If f(x) = x + 2 , then f’ (f(x)) at x= 4 is
(1) 8
(2) 1
(3) 4
(4) 5
Answer:
(2) 1

Explaination:
f(x) x + 2
f’ (f(x)) = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (f(x))
= \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x + 2) = 1

Question 15.
If y = \(\frac{(1-x)^{2}}{x^{2}}\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) is
(1) \(\frac{2}{x^{2}}+\frac{2}{x^{3}}\)
(2) \(-\frac{2}{x^{2}}+\frac{2}{x^{3}}\)
(3) \(-\frac{2}{x^{2}}-\frac{2}{x^{3}}\)
(4) \(-\frac{2}{x^{3}}+\frac{2}{x^{2}}\)
Answer:
(4) y = \(\frac{(1-x)^{2}}{x^{2}}\)

Explaination:

Question 16.
If pv = 81, then \(\frac{\mathbf{d} \mathbf{p}}{\mathbf{d v}}\) at v = 9 is
(1) 1
(2) – 1
(3) 2
(4) – 3
Answer:
(2) – 1

Explaination:

Question 17.

(1) 0
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

Question 18.
It is given that f’ (a) exists, then

(1) f(a) – af'(a)
(2) f'(a)
(3) – f'(a)
(4) f(a) + af'(a)
Answer:
(1) f(a) – af'(a)

Explaination:

Question 19.
If f(x) = then f’ (2) is
(1) 0
(2) 1
(3) 2
(4) does not exist
Answer:
(3) 2

Explaination:

Question 20.
If g(x) = (x² + 2x + 3) f(x) and f(0) – 5 and then g'(0) is
(1) 20
(2) 14
(3) 18
(4) 12
Answer:

Question 21.
If f(x) = then at x = 3, f'(x) is
(1) 1
(2) – 1
(3) 0
(4) does not exist
Answer:
(2) – 1

Explaination:

From equations (1) and (2) we get
f’ (3^–) ≠ f’ (3⁺)
∴ limit of f(x) does not exist at x = 3
f’ (x) does not exist at x = 3

Question 22.
The derivative of f(x)= x|x| at x = – 3 is
(1) 6
(2) – 6
(3) does not exist
(4) 0
Answer:
(1) 6

Explaination:
f(x) = x|x|
f(x) = x(-x) ⇒ f(x) = – x²
f ‘(x) = -(2x)
f ‘(-3) = -(2) (-3) = 6

Question 23.
If f(x) = then which one of the following is true?
(1) f(x) is not differentiable at x = a
(2) f(x) is discontinuous at x = a
(3) f(x) is continuous for all x in R
(4) f(x) is differentiable for all x ≥ a
Answer:
(1) f(x) is not differentiable at x = a

Explaination:


f’ (a⁺) = 3 ………. (2)
From equations (1) and (2) we get
f'(a^–) ≠ f'(a⁺)
∴ f’ (x) does not exist at x = a
∴ f(x) is not differentiable at x = a

Question 24.
If f(x) = is differentiable at x = 1, then
(1) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(2) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(3) a = \(-\frac{1}{2}\), b = \(-\frac{3}{2}\)
(4) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Answer:
(3) a = \(-\frac{1}{2}\), b = \(-\frac{3}{2}\)

Explaination:

Question 25.
The number of points in R in which the function f(x) = |x – 1| + |x – 3| + sin x is not differentiable, is
(1) 3
(2) 2
(3) 1
(4) 4
Answer:
(2) 2

Explaination:
f(x) = |x – 1| + |x – 3| + sin x is not differentiable at x = 1, and x = 3

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1

Integrate the following with respect to x:

Question 1.
(i) x¹¹
(ii) \(\frac{1}{x^{7}}\)
(iii) \(\sqrt[3]{x^{4}}\)
(iv) (x⁵)^1/8
Answer:
(i) x¹¹

(ii) \(\frac{1}{x^{7}}\)

(iii) \(\sqrt[3]{x^{4}}\)

(iv) (x⁵)^1/8

Question 2.
(i) \(\frac{1}{\sin ^{2} x}\)
(ii) \(\frac{\tan x}{\cos x}\)
(iii) \(\frac{\cos x}{\sin ^{2} x}\)
(iv) \(\frac{1}{\cos ^{2} x}\)
Answer:
(i) \(\frac{1}{\sin ^{2} x}\)
\(\frac{1}{\sin ^{2} x}\) = ∫cosec² x dx
= – cot x + c

(ii) \(\frac{\tan x}{\cos x}\)
\(\frac{\tan x}{\cos x}\) = ∫sec x tan x dx
= sec x + c

(iii) \(\frac{\cos x}{\sin ^{2} x}\)

(iv) \(\frac{1}{\cos ^{2} x}\)
\(\frac{1}{\cos ^{2} x}\) = ∫sec² x dx
= tan x + c

Question 3.
(i) 12³
(ii) \(\frac{x^{24}}{x^{25}}\)
(iii) e^x
Answer:
(i) 12³
∫12³ dx = 12³ ∫dx = 12³x + c

(ii) \(\frac{x^{24}}{x^{25}}\)

(iii) e^x
∫e^x dx = e^x + c

Question 4.
(i) (1 + x²)^-1
(ii) (1 – x²)^-1/2
Answer:
(i) (1 + x²)^-1

(ii) (1 – x²)^-1/2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2

Integrate the following functions with respect to x

Question 1.
(i) (x + 5)⁶
(ii) \(\frac{1}{(2-3 x)^{4}}\)
(iii) \(\sqrt{3 x+2}\)
Answer:
(i) (x + 5)⁶

(ii) \(\frac{1}{(2-3 x)^{4}}\)

(iii) \(\sqrt{3 x+2}\)

Question 2.
(i) sin 3x
(ii) cos (5 – 11x)
(iii) cosec² (5x – 7)
Answer:
(i) sin 3x
∫sin (ax + b) dx = – \(\frac{1}{a}\) cos (ax + b) + c
∫sin 3x dx = – \(\frac{1}{3}\) cos 3x + c

(ii) cos (5 – 11x)

(iii) cosec² (5x – 7)
[∫cosec² (ax + b) dx = – \(\frac{1}{a}\) cot (ax + b) + c]
∫cosec² (5x – 7) dx = – \(\frac{1}{5}\) cot (5x – 7) + c

Question 3.
(i) e^{3x – 6}
(ii) e^{8 – 7x}
(iii) \(\frac{1}{6-4 x}\)
Answer:
(i) e^{3x – 6}
[∫e^ax+b dx = \(\frac{1}{a}\) e^{ax + b} + c]
∫e^{3x – 6} dx = \(\frac{1}{3}\) e^{3x – 6} + c

(ii) e^{8 – 7x}
[∫e^ax+b dx = \(\frac{1}{a}\) e^{ax + b} + c]
∫e^{8 – 7x} = \(\frac{1}{-7}\) e^{8 – 7x} + c
∫e^{8 – 7x} = \(-\frac{1}{7}\) e^{8 – 7x} + c

(iii) \(\frac{1}{6-4 x}\)

Question 4.
(i) sec² \(\frac{x}{5}\)
(ii) cosec (5x + 3) cot (5x + 3)
(iii) 30 sec (2 – 15x) tan (2 – 15x)
Answer:
(i) sec² \(\frac{x}{5}\)
[∫sec² (ax + b)dx = \(\frac{1}{a}\) tan (ax + b) + c]

(ii) cosec (5x + 3) cot (5x + 3)
[∫cosec (ax + b) cot (ax + b) dx = – \(\frac{1}{a}\) cosec (ax + b) + c]
∫cosec (5x + 3) cot (5x + 3) dx = – \(\frac{1}{5}\) cosec (5x + 3) + c]

(iii) 30 sec (2 – 15x) tan (2 – 15x)
[∫sec (ax + b) tan (ax + b)dx = \(\frac{1}{a}\) sec (ax + b) + c]
∫30 sec (2 – 15x) tan (2 – 15x)dx = 30 × \(\frac{1}{-15}\) × sec (2 – 15x) + c
= – 2 sec (2 – 15x) + c

Question 5.
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)
(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)
(iii) \(\frac{1}{1+36 x^{2}}\)
Answer:
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)

(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)

(iii) \(\frac{1}{1+36 x^{2}}\)