Question

${\int }_0^{\frac{\pi }{4}}\frac{{\cos }^{4}x-{\sin }^{4}x}{\sqrt{1+\sin 2x}}dx=\sqrt{2}$

• A. True
• B. False

Answer 1

The correct option is B False

We have,

$I={\int }_0^{\frac{\pi }{4}}\frac{{\cos }^{4}x-{\sin }^{4}x}{\sqrt{1+\sin 2x}}dx$

$I={\int }_0^{\frac{\pi }{4}}\frac{({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)}{\sqrt{1+\sin 2x}}dx$

$I={\int }_0^{\frac{\pi }{4}}\frac{({\cos }^{2}x-{\sin }^{2}x)}{\sqrt{1+\sin 2x}}dx$ $\because {\sin }^{2}x+{\cos }^{2}x=1$

$I={\int }_0^{\frac{\pi }{4}}\frac{\cos 2x}{\sqrt{1+\sin 2x}}dx$ $\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x$

Let

$t=1+\sin 2x$

$\frac{dt}{dx}=0+2\cos 2x$

$\frac{dt}{2}=\cos 2xdx$

Therefore,

$I=\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{t}}$

$I=\frac{1}{2}\times 2[\sqrt{t}{]}_1^2$

$I=\sqrt{2}-\sqrt{1}$

$I=\sqrt{2}-1$

Hence, this is the answer.