Question

If $I_{(n,m)}=\int \frac{{\sin }^{n}x}{{\cos }^{m}x}dx,$ then the value of $I_{(3,4)}=$
(where $C$ is integration constant)

• A. $\frac{1}{3}\frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\sec x+C$
• B. $\frac{1}{4}{\tan }^{2}x\cdot {\sec }^{2}x-\frac{2}{3}\sec x+C$
• C. $\frac{2}{3}\frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{1}{3}\sec x+C$
• D. $\frac{1}{4}{\tan }^{2}x\cdot \sec x-\frac{2}{3}\sec x+C$

Answer 1

The correct option is A $\frac{1}{3}\frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\sec x+C$

Using Integration by parts for $I_{(n,m)}$, we get
$I_{(n,m)}=\int {\sin }^{n-1}x\frac{\sin x}{{\cos }^{m}x}dx={\sin }^{n-1}x\cdot \frac{(\cos x{)}^{-m+1}}{(m-1)}-\int (n-1){\sin }^{n-2}x\cdot \cos x\cdot \frac{(\cos x{)}^{-m+1}}{(m-1)}dx=\frac{1}{m-1}\cdot \frac{{\sin }^{n-1}x}{{\cos }^{m-1}x}-\frac{(n-1)}{(m-1)}\cdot \int \frac{{\sin }^{n-2}x}{{\cos }^{m-2}x}dx{I}_{n,m}=\frac{1}{(m-1)}\cdot \frac{{\sin }^{n-1}x}{{\cos }^{m-1}x}-\frac{(n-1)}{(m-1)}\cdot I_{(n-2,m-2)}$ is the required reduction formula.
$\Rightarrow I_{(3,4)}=\frac{1}{3}\cdot \frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\cdot I_{(1,2)}$
$\Rightarrow I_{(3,4)}=\frac{1}{3}\cdot \frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\int \frac{\sin x}{{\cos }^{2}x}dx=\frac{1}{3}\cdot \frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\int \sec x\cdot \tan xdx=\frac{1}{3}\frac{{\sin }^{2}x}{{\cos }^{3}x}-\frac{2}{3}\sec x+C$