Question

$\int si{n}^3\left(x\right).co{s}^5\left(x\right)dx$

• A. $\frac{co{s}^6\left(x\right)}{4}-\frac{co{s}^6\left(x\right)}{6}+C$
• B. $\frac{co{s}^8\left(x\right)}{8}-\frac{co{s}^6\left(x\right)}{6}+C$
• C. $\frac{co{s}^4\left(x\right)}{6}-\frac{co{s}^6\left(x\right)}{6}+C$
• D. $\frac{co{s}^8\left(x\right)}{8}-\frac{co{s}^6\left(x\right)}{3}+C$

Answer 1

The correct option is B

$\frac{co{s}^8\left(x\right)}{8}-\frac{co{s}^6\left(x\right)}{6}+C$

We can see that this is the Integration of type $\int si{n}^{m}x.co{s}^{n}xdx$. Out of all the cases discussed, we can see that it is the case where m & n both are odd. So we can substitute either sin(x) or cos(x) equal to t.

Let’s substitute cos(x) = t

So, - sin(x) dx = dt

$\int si{n}^3\left(x\right).co{s}^5\left(x\right)dx=\int si{n}^2\left(x\right).co{s}^5\left(x\right).sin\left(x\right)dxOr-\int (1-t^2).t^5.dt\int (t^2-1).t^5.dt\int (t^7-t^5).dt=\frac{t^8}{8}-\frac{t^6}{6}+c$

Substitute t = cosx

$=\frac{co{s}^8\left(x\right)}{8}-\frac{co{s}^6\left(x\right)}{6}+c$