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) \(\int_{-\pi / 2}^{\pi / 2}\) (x^{5} + x cos x + tan³ x) dx

Solution:

\(\int_{-\pi / 2}^{\pi / 2}\) (x^{5} + x cos x + tan³ x) dx

= \(\int_{-\pi / 2}^{\pi / 2}\) (x^{5} + x cos x + tan³ x) dx + \(\int_{-\pi / 2}^{\pi / 2}\)

Let f(x) = x^{5} + x cos x + tan³x

f(-x) = -x^{5} – 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) \(\int_{0}^{π}\) sin^{4} x cos³ x dx

Solution:

\(\int_{0}^{π}\) sin^{4} x cos³ x dx

f(x) = sin^{4}x cos³x

f(2π – x) = sin^{4}(2π – x) cos³ (2π – x)

= sin^{4}x cos³x

f(2π – x) = f(x)

Limit from 0 to π tends to 0 to 0

∴ Integral value = 0

∴ \(\int_{0}^{π}\) sin^{4} 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:

I_{1} = \(\int_{0}^{sin^2x}\) 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

I_{1} = \(\int_{0}^{cos^2x}\) 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)