Question

$\int {\sin }^{3}x.{\cos }^{2}xdx=$

• A. $\left(\frac{{\sin }^{5}x}{5}\right)-\left(\frac{{\sin }^{3}x}{3}\right)+c$
• B. $\left(\frac{{\sin }^{5}x}{5}\right)+\left(\frac{{\sin }^{3}x}{3}\right)+c$
• C. $\left(\frac{{\cos }^{5}x}{5}\right)-\left(\frac{{\cos }^{3}x}{3}\right)+c$
• D. $\left(\frac{{\cos }^{5}x}{5}\right)+\left(\frac{{\cos }^{3}x}{3}\right)+c$

Answer 1

The correct option is C

$\left(\frac{{\cos }^{5}x}{5}\right)-\left(\frac{{\cos }^{3}x}{3}\right)+c$

Explanation For The Correct Option:

Evaluating the integral:

$\int {\sin }^{3}x.{\cos }^{2}xdx$

$$\begin{gathered} =\int {\sin }^{3}x\left(1-{\sin }^{2}x\right)dx\left[\because {\cos }^{2}x=1-{\sin }^{2}x\right] \\ =\int \left({\sin }^{3}x-{\sin }^{5}x\right)dx \\ =\int {\sin }^{3}xdx-\int {\sin }^{5}xdx \\ =\int {\sin }^{2}x\sin xdx-\int {\sin }^{4}x\sin xdx \\ =\int \left(1-{\cos }^{2}x\right)\sin xdx-\int {\left(1-{\cos }^{2}x\right)}^2\sin xdx\left[\because {\sin }^{2}x=1-{\cos }^{2}x\right] \end{gathered}$$

Substituting $t=\cos x$ and substituting the value of $dx$ after differentiating $t$ w.r.t $x$

$dt=-\sin xdx$

$$\begin{gathered} =-\int \left(1-t^2\right)dt+\int {\left(1-t^2\right)}^{2}dt \\ =-\int \left(1-t^2\right)dt+\int \left(1+t^4-2t^2\right)dt \\ =-\int dt+\int t^{2}dt+\int dt+\int t^{4}dt-2\int t^{2}dt \\ =-\frac{t^3}{3}+\frac{t^5}{5}+C[c\text{is an integrating constant}] \\ =\left(\frac{{\cos }^{5}x}{5}\right)-\left(\frac{{\cos }^{3}x}{3}\right)+c\left[\text{substituting}t=\cos x\right] \end{gathered}$$

Therefore, $\int {\sin }^{3}x.{\cos }^{2}xdx=$$\left(\frac{{\cos }^{5}x}{5}\right)-\left(\frac{{\cos }^{3}x}{3}\right)+c$

Hence, option (C) is the correct answer.