Question

${\int }_0^{{\pi }_{/}2}\frac{{\cos }^{3/2}x}{{\cos }^{3/2}x+{\sin }^{3/2}x}dx=$

• A. $\pi /3$
• B. $\pi /2$
• C. $\pi /4$
• D. $\pi /8$

Answer 1

The correct option is C $\pi /4$

Let $I={\int }_0^{{\pi }_{/}2}\frac{{\cos }^{3/2}x}{{\cos }^{3/2}x+{\sin }^{3/2}x}dx$ ......(1)

Now using, ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$I={\int }_0^{{\pi }_{/}2}\frac{{\cos }^{3/2}(\frac{\pi }{2}-x)}{{\cos }^{3/2}(\frac{\pi }{2}-x)+{\sin }^{3/2}(\frac{\pi }{2}-x)}dx={\int }_0^{{\pi }_{/}2}\frac{{\sin }^{3/2}x}{{\sin }^{3/2}x+{\cos }^{3/2}x}dx$ ......(2)

Now, $\left(1\right)+\left(2\right)\Rightarrow 2I={\int }_0^{\pi /2}1\cdot dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$