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

Category: QuestionsFind \(\int_0^{\frac{\sqrt{π}}{2}} 2x \,cos⁡ x^2 \,dx\).
Editor">Editor Staff asked 11 months ago

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
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1 Answers
Editor">Editor Staff answered 11 months ago

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 \(x={\frac{\sqrt{π}}{2}}, t=\frac{π}{4}\)
 
∴\(\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx=\int_0^{\frac{π}{4}} \,cos⁡t \,dt\)
 
\(I =[sin⁡t]_0^{\frac{π}{4}}=sin⁡ \frac{π}{4}-sin⁡0=1/\sqrt{2}\).