Question

If the relation between the order of integrals of sin(x) can be given by
$\int si{n}^n\left(x\right)dx=\frac{-si{n}^{n-1}\left(x\right).cos\left(x\right)}{n}+\frac{n-1}{n}\int si{n}^{n-2}\left(x\right)dx$; n > 0
Then find $\int si{n}^3\left(x\right)dx.$

• A. $\frac{-cos\left(x\right)}{3}(1+si{n}^2(x\left)\right)+C$
• B. $\frac{-cos\left(x\right)}{5}(1+si{n}^2(x\left)\right)+C$
• C. $\frac{-sin\left(x\right)}{3}(1+si{n}^2(x\left)\right)+C$
• D. $\frac{-sin\left(x\right)}{5}(1+si{n}^2(x\left)\right)+C$

Answer 1

The correct option is A $\frac{-cos\left(x\right)}{3}(1+si{n}^2(x\left)\right)+C$

The given relation is nothing but the reduction formula for sin(x). We’ll use the the above formula for n = 3.
So, $\int si{n}^3\left(x\right)dx=\frac{-si{n}^2\left(x\right).cos\left(x\right)}{3}+\frac{1}{3}\int sin\left(x\right)dx$
$\int si{n}^3\left(x\right)dx=\frac{-si{n}^2\left(x\right).cos\left(x\right)}{3}-\frac{1}{3}cos\left(x\right)+C(Using\int sin\left(x\right)dx=-cos\left(x\right))$
Or $\frac{-cos\left(x\right)}{3}(1+si{n}^2(x\left)\right)+C$