1.

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»

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

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

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

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


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