Question

Find \(\int_0^{π/4} \frac{5 \,sin⁡(tan^{-1}⁡x)}{1+x^2} \,dx\).

(a) 5-\(\frac{1}{\sqrt{2}}\)

(b) 5+\(\frac{5}{\sqrt{2}}\)

(c) -5+\(\frac{5}{\sqrt{2}}\)

(d) 5-\(\frac{5}{\sqrt{2}}\)

This question was addressed to me in semester exam.

I'd like to ask this question from Evaluation of Definite Integrals by Substitution topic in portion Integrals of Mathematics – Class 12

Answer 1

Right answer is (d) 5-\(\frac{5}{\sqrt{2}}\)

Easy explanation: I=\(\int_0^1 \frac{5 \,sin⁡(tan^{-1)}x}{1+x^2} \,dx\)

Let tan^-1⁡x=t

Differentiating w.r.t x, we get

\(\frac{1}{1+x^2} \,dx=dt\)

The new limits

When x=0, t=tan^-1⁡0=0

When x=1, t=tan^-1)1=π/4

∴\(\int_0^1 \frac{5 \,sin⁡(tan^{-1}⁡x)}{1+x^2} \,dx=\int_0^{π/4} \,5 \,sin⁡t \,dt\)

=\(5[-cos⁡t]_0^{π/4}=-5[cos⁡t]_0^{π/4}\)

\(=-5(cos⁡ \frac{π}{4}-cos⁡0)=-5(\frac{1}{\sqrt{2}-1})=5-\frac{5}{\sqrt{2}}\)