Question

Evaluate$\int {\sin }^{4}x{\cos }^{3}xdx.$

Answer 1

Hint: Use substitution method i.e. substitute $\sin x=t$ for easy simplification.

Let, $I=\int {\sin }^{4}x{\cos }^{3}xdx=\int {\sin }^{4}x{\cos }^{2}x\cos xdx$
As we know${\cos }^{2}x=\left(1-{\sin }^{2}x\right)$, so substitute this value.
$\Rightarrow I=\int {\sin }^{4}x{\cos }^{3}xdx=\int {\sin }^{4}x\left(1-{\sin }^{2}x\right)\cos xdx.................\left(1\right)$
Now, let $\sin x=t$
Differentiate above equation w.r.t.$x$
As we know $\sin x$differentiation is $\cos x$
$\Rightarrow \cos xdx=dt$
So, substitute this value in equation (1).
$$\begin{gathered} I=\int {\sin }^{4}x\left(1-{\sin }^{2}x\right)\cos xdx=\int t^4\left(1-t^2\right)dt \\ \Rightarrow I=\int \left(t^4-t^6\right)dt \end{gathered}$$
Now, integrate it, as we know$\int t^{n}dt=\left[{\displaystyle \frac{t^{n+1}}{n+1}}\right]$, so apply this property.
$\Rightarrow I=\left[{\displaystyle \frac{t^5}{5}}-{\displaystyle \frac{t^7}{7}}\right]+c$, (where c is some arbitrary integration constant)
L.C.M of 5 and 7 is 35, and take $t^5$as common
$\Rightarrow I={\displaystyle \frac{t^5}{35}}\left[7-5t^2\right]+c$
Now re-substitute the value of $t=\sin x$
$\Rightarrow I={\displaystyle \frac{\left({\sin }^{5}x\right)}{35}}\left[7-5\left({\sin }^{2}x\right)\right]+c$
So, this is the required value of the integration.

Note: In such types of questions always choose substitution which makes integration simple, in above integration we choose $\sin x=t$, so it makes integration simple, then we easily integrate using some basic property of integration which is stated above, then simplify we will get the required answer.