$\Rightarrow dx = a{\sec ^2}\theta \,d\theta ,$ then we have
$I = \int_0^{\pi /4} {\frac{{{a^4}{{\tan }^4}\theta .\,a{{\sec }^2}\theta \,d\theta }}{{{a^8}{{\sec }^8}\theta }}} $
==> $\frac{1}{{{a^3}}}\int_0^{\pi /4} {{{\sin }^4}\theta } {\cos ^2}\theta \,d\theta $
$= I = \frac{1}{{{a^3}}}\left[ {\int_0^{\pi /4} {({{\sin }^4}\theta } - {{\sin }^6}\theta } \right]\,d\theta $
$ = \frac{1}{{{a^3}}}\int_0^{\pi /4} {\left[ {\frac{{{{(1 - \cos 2\theta )}^2}}}{4} - \frac{{{{(1 - \cos 2\theta )}^3}}}{8}} \right]\,} d\theta $
$ = \frac{1}{{8{a^3}}}\int_0^{\pi /4} {{\rm{ }}(1 + \cos 2\theta } )(1 + {\cos ^2}2\theta - 2\cos 2\theta )d\theta $
$ = \frac{1}{{8{a^3}}}\int_0^{\pi /4} {(1 - \cos 2\theta - {{\cos }^2}2\theta + {{\cos }^3}2\theta )\,d\theta } $
$ = \frac{1}{{32{a^3}}}\int_0^{\pi /4} {(2 - \cos 2\theta - 2\cos 4\theta + \cos 6\theta )d\theta } $
$ = \frac{1}{{32{a^3}}}\left[ {2\theta - \frac{{\sin 2\theta }}{2} - \frac{{\sin 4\theta }}{2} + \frac{{\sin 6\theta }}{6}} \right]_0^{\pi /4}$
$ = \frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Match List $I$ with List $II$ and select the correct answer using the code given below the lists :
| List $I$ | List $II$ |
| $P.$ $\quad$m= | $1.$ $\quad\frac{1}{2}$ |
| $Q.$ $\quad$Maximum area of $\triangle E F G$ is | $2.$ $\quad4$ |
| $R.$ $\quad y_0=$ | $3.$ $\quad2$ |
| $S.$ $\quad y_1=$ | $4.$ $\quad1$ |
Codes: $ \quad P \quad Q \quad R \quad S $