Question

$\int {\sin }^{3}x{\cos }^{4}xdx$ is equal to.

• A.
  $\frac{1}{7}{\cos }^{7}x+\frac{1}{5}{\cos }^{5}x+C$
• B.
  $\frac{1}{7}{\cos }^{7}x-\frac{1}{5}{\cos }^{5}x+C$
• C.
  $-\frac{1}{7}{\cos }^{7}x+\frac{1}{5}{\cos }^{5}x+C$
• D.
  $-\frac{1}{7}{\cos }^{7}x-\frac{1}{5}{\cos }^{5}x+C$

Answer 1

The correct option is B

$\frac{1}{7}{\cos }^{7}x-\frac{1}{5}{\cos }^{5}x+C$
For solving these types of integrals we check the powers on $\mathrm{cosine}\mathrm{and}\mathrm{sine}\mathrm{term}$ Here, the power on $\mathrm{sine}$ term is odd, so we make a substitution $u=\cos x$
then we get, $du=-\sin xdx$
Then our integral becomes, $I=-\int u^4(1-u^2)du\Rightarrow I=-\int (u^4-u^6)du\Rightarrow I=-\frac{u^5}{5}+\frac{u^7}{7}+C\mathrm{Subsituting}\mathrm{back}u=\cos x\mathrm{we}\mathrm{get}:I=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+C$
Thus, Option b. is correct.