#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 1

Answer: -     $\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C$

Hint: - Use substitution method to solve this integral

Given:-$\int \sin ^{4} x \cos ^{3} x d x$

Solution: - Let, $I=\int \sin ^{4} x \cos ^{3} x d x$

The exponent of $\cos x$  is odd, so we substitute $\sin x=t\Rightarrow \cos dt$then,
$I=\int t^{4} \cos ^{3} x \frac{d t}{\cos x} \quad\quad\quad\quad\quad\quad\quad[\because \sin x=t]$
$=\int t^{4} \cos ^{2} x d t$
$\begin{array}{ll} =\int t^{4}\left(1-\sin ^{2} x\right) d t &\quad\quad\quad\quad\quad\quad {\left[\because \sin ^{2} x=\cos ^{2} x=1\right]} \\ \\=\int t^{4}\left(1-t^{2}\right) d t & \quad\quad\quad\quad\quad\quad{[\because \sin x=t]} \end{array}$
\begin{aligned}\\ &=\int\left(t^{4}-t^{4} t^{2}\right) d t \\ &=\int\left(t^{4}-t^{6}\right) d t \\ &=\int t^{4} d t-\int t^{6} d t \end{aligned}
$=\frac{t^{4+1}}{4+1}-\frac{t^{6+1}}{6+1}+C$                                  $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right]$
$=\frac{t^{5}}{5}-\frac{t^{7}}{7}+C$

$=\frac{\sin ^{5} x}{5}-\frac{\sin ^{7}}{7}+C$                                      $\left [ \because t=\sin x \right ]$