Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3

Question 1.
Evaluate the following definite integrals.
(i) $$\int_{3}^{4}$$ $$\frac{dx}{x^2-4}$$
Solution:

(ii) $$\int_{-1}^{1}$$ $$\frac{dx}{x^2+2x+5}$$
Solution:

(iii) $$\int_{0}^{1}$$ $$\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ dx
Solution:

(iv) $$\int_{0}^{\pi / 2}$$ ($$\frac{1+sin}{1+cosx}$$)dx
Solution:

(v) $$\int_{0}^{\pi / 2}$$ $$\sqrt{cos θ}$$ sin³θ dθ
Solution:

(vi) $$\int_{0}^{1}$$ $$\frac{1-x^2}{(1+x^2)^2}$$ dx
Solution:

Question 2.
Evaluate the following integrals using properties of integration:
(i) $$\int_{-5}^{5}$$ x cos ($$\frac{e^x-1}{e^x+1}$$) dx
Solution:

(ii) $$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x cos x + tan³ x) dx
Solution:
$$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x cos x + tan³ x) dx
= $$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x cos x + tan³ x) dx + $$\int_{-\pi / 2}^{\pi / 2}$$
Let f(x) = x5 + x cos x + tan³x
f(-x) = -x5 – x cos x – tan³x
f(x) = -f(-x)
f(x) is an odd function

(iii) $$\int_{-\pi / 4}^{\pi / 4}$$ sin² x dx
Solution:
I = $$\int_{-\pi / 4}^{\pi / 4}$$ sin² x dx
f(x) = sin²x
f(-x) = sin²(-x) = sin²x
f(x) = f(-x)
f(x) is an even function

(iv) $$\int_{0}^{2π}$$ x log($$\frac{3+cos x}{3-cos x}$$)dx
Solution:

(v) $$\int_{0}^{π}$$ sin4 x cos³ x dx
Solution:
$$\int_{0}^{π}$$ sin4 x cos³ x dx
f(x) = sin4x cos³x
f(2π – x) = sin4(2π – x) cos³ (2π – x)
= sin4x cos³x
f(2π – x) = f(x)

Limit from 0 to π tends to 0 to 0
∴ Integral value = 0
∴ $$\int_{0}^{π}$$ sin4 x cos³ x dx = 0

(vi) $$\int_{0}^{1}$$ |5x – 3|dx
Solution:

(vii) $$\int_{0}^{sin^2x}$$ sin-1 √t dt + $$\int_{0}^{cos^2x}$$ cos-1 √t dt
Solution:
I1 = $$\int_{0}^{sin^2x}$$ sin-1 √t dt
Put sin-1 √t = θ
√t = sin θ
$$\frac{1}{2√t}$$ dt = cos θ dθ
dt = 2√t cos θ dθ
= 2 sin θ cos θ dθ
dt = sin 2θ dθ

I1 = $$\int_{0}^{cos^2x}$$ cos-1 √t dt
Put cos-1 √t = θ
√t = cos θ
$$\frac{1}{2√t}$$ dt = -sin θ dθ
dt = -2√t sin θ dθ
= -2 cos θ sin θ dθ
dt = -sin 2θ dθ

(viii) $$\int_{0}^{1}$$ $$\frac{log(1+x)}{1+x^2}$$ dx
Solution:

(ix) $$\int_{0}^{π}$$ $$\frac{x sin x}{1+sin x}$$ dx
Solution:

(x) $$\int_{π/8}^{3π/8}$$ $$\frac{1}{1+\sqrt{tan x}}$$ dx
Solution:

(xi) $$\int_{0}^{π}$$ x[sin²(sin x) + cos² (cos x)] dx
Solution:
Let I = $$\int_{0}^{π}$$ x[sin²(sin x) + cos² (cos x)] dx
f(x) = sin² (sin x) + cos² (cos x)
f(π – x) = sin² (sin π – x)) + cos² (cos(π – x))
= sin² (sin x) + cos² (cos x)
f(x) = f(π – x)