Question

Find \(\int_0^{\frac{\sqrt{π}}{2}} 2x \,cos⁡ x^2 \,dx\).

(a) 1

(b) \(\frac{1}{\sqrt{2}}\)

(c) –\(\frac{1}{\sqrt{2}}\)

(d) \(\sqrt{2}\)

This question was posed to me in unit test.

This interesting question is from Evaluation of Definite Integrals by Substitution in section Integrals of Mathematics – Class 12

Answer 1

Right choice is (b) \(\frac{1}{\sqrt{2}}\)

The explanation is: I=\(\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx\)

Let x^2=t

Differentiating w.r.t x, we get

2x dx=dt

The new limits

When x=0,t=0

When \(x={\frac{\sqrt{π}}{2}}, t=\frac{π}{4}\)

∴\(\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx=\int_0^{\frac{π}{4}} \,cos⁡t \,dt\)

\(I =[sin⁡t]_0^{\frac{π}{4}}=sin⁡ \frac{π}{4}-sin⁡0=1/\sqrt{2}\).