Question

If$I_1={\int }_0^{\frac{\mathrm{\pi }}{4}}{\sin }^{2}xdx$and$I_2={\int }_0^{\frac{\mathrm{\pi }}{4}}{\cos }^{2}xdx$, then

• A. $I_1=I_2$
• B. $I_1<I_2$
• C. $I_1>I_2$
• D. $I_2=I_1+\pi /4$

Answer 1

The correct option is B

$I_1<I_2$

Explanation for correct option:

Find the relation:

Taking $I_1$first,

$\begin{array}{c}{I}_1={\int }_0^{\frac{\pi }{4}}{\sin }^{2}xdx\\ =\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(1-\cos 2x)dx\mathbf{[}\mathbf{\because }\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{]}\\ =\frac{1}{2}{\left[x-\frac{\sin 2x}{2}\right]}_0^{\frac{\pi }{4}}\\ =\frac{1}{2}\left[(\frac{\pi }{4}-\frac{\sin \frac{\pi }{2}}{2})-0\right]\\ =\frac{1}{2}\left[\frac{\pi }{4}-\frac{1}{2}\right]\dots \left(1\right)\end{array}$

$\begin{array}{l}{I}_2={\int }_0^{\frac{\pi }{4}}{\cos }^{2}xdx\\ =\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(1+\cos 2x)dx\left[\because 1+\cos 2x=2{\cos }^{2}x\right]\\ =\frac{1}{2}{\left[x+\frac{\sin 2x}{2}\right]}_0^{\frac{\pi }{4}}\\ =\frac{1}{2}\left[\frac{\pi }{4}+\frac{1}{2}\right]\dots \left(2\right)\\ \end{array}$

From (1) and (2) it is clear that,

$I_2>I_1$

Hence, the correct option is B.