Let I = ^7 x d x = ( ^2 x )^3 x d x = (1- ^2 x )^3 x d x Put x=t x d d = dt I = (1- t ^2 )^3 dt = (1-3 t ^2+3 t ^4- t ^6 ) dt = 1 dt -3 t ^2 dt +3 t ^4 dt - t ^6 dt .= t -3 ( t ^3 3 )+3 t ^5 5 )- t ^7 7 + c I = x- ^3 x+3 5 ^5 x-1 7 ^7 x+ c

^7 x d x

Question

$\int \cos ^7 x d x$

✓

Answer

$ \text { Let } I =\int \cos ^7 x d x$
$=\int\left(\cos ^2 x\right)^3 \cdot \cos x d x$
$=\int\left(1-\sin ^2 x\right)^3 \cdot \cos x d x $
Put $\sin x=t$
$ \therefore \cos x d d = dt$
$\therefore I =\int\left(1- t ^2\right)^3 dt$
$=\int\left(1-3 t ^2+3 t ^4- t ^6\right) dt$
$=\int 1 \cdot dt -3 \int t ^2 dt +3 \int t ^4 dt -\int t ^6 dt$
$\left.= t -3\left(\frac{ t ^3}{3}\right)+3 \frac{ t ^5}{5}\right)-\frac{ t ^7}{7}+ c$
$\therefore I =\sin x-\sin ^3 x+\frac{3}{5} \sin ^5 x-\frac{1}{7} \sin ^7 x+ c $

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