Question

$\int si{n}^{3}x.co{s}^{4}xdx$

• A. $\frac{co{s}^7\left(x\right)}{7}-\frac{co{s}^5\left(x\right)}{5}+c$
• B. $\frac{co{s}^7\left(x\right)}{5}+\frac{co{s}^5\left(x\right)}{7}+c$
• C. $\frac{co{s}^7\left(x\right)}{3}+\frac{co{s}^5\left(x\right)}{5}+c$
• D. $\frac{co{s}^7\left(x\right)}{5}+\frac{co{s}^5\left(x\right)}{3}+c$

Answer 1

The correct option is A

$\frac{co{s}^7\left(x\right)}{7}-\frac{co{s}^5\left(x\right)}{5}+c$

We can see here that the above expression is in $\int si{n}^{m}x.co{s}^{n}xdx$ form with having one power odd and another even. We saw that in such case we substitute the one which has even power.

Since cos(x) has even power we’ll substitute cos (x) = t .

$\int si{n}^{3}x.co{s}^{4}xdx$

$Or\int si{n}^2\left(x\right).sin\left(x\right).co{s}^4\left(x\right)dx$

Let’s substitute cos(x) = t

- sin(x). dx= dt

So, we’ll have -

$-\int (1-t^2).t^{4}dt$

$Or-\int t^{4}dt+\int t^{6}dt$

$\frac{t^7}{7}-\frac{t^5}{5}+c$

Let’s substitute t = cos(x)

And we’ll have

$\frac{co{s}^7\left(x\right)}{7}-\frac{co{s}^5\left(x\right)}{5}+c$