Q113 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
If $I_m=\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} I_{m-2}$, then $g(x)=$
- A$\frac{x^m \sin n x}{n}+\frac{m(m-1) x^{m-1} \cos n x}{n^2}$
- B$\frac{x^m \cos n x}{n}+\frac{x^{m-1} m(m-1)}{n^2} \sin n x$
- C$\frac{m}{n} \sin n x+\frac{m}{n^2} x^{m-1} \cos n x$
- Correct answerD$\frac{x^m \sin n x}{n}+\frac{m}{n^2} x^{m-1} \cos n x$
Solution
Given that,
$I_m=\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} I_{m-2}$ $\ldots(i)$
$I_{m-2}=\int x^{m-2} \cos n x d x$
Solving integration by parts
$\left.=\cos n x \int x^{m-2} d x-\int\left(\frac{d}{d x} \cos n x\right) \int x^{m-2} d x\right) d x$
$=\cos n x\left[\frac{x^{m-1}}{m-1}\right]+n \int \sin n x \frac{x^{m-1}}{m-1} d x$
$I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \int x^{m-1} \sin n x d x$
Again integrating by parts method,
$I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1}$
$\left\{\sin n x \int x^{m-1} d x-\int\left(\frac{d}{d x} \sin n x \int x^{m-1} d x\right) d x\right\}$
$=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1}$
$\left\{\frac{\sin n x \cdot x^m}{m}-\int n \cdot \cos n x \frac{x^m}{m} d x\right\}$
$=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \cdot \frac{1}{m}$
$\left\{x^m \sin n x-n \int x^m \cos n x d x\right\}$
$=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)}\left(x^m \sin n x-n I_m\right)$
$I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x-\frac{n^2}{m(m-1)} I_m$
or $\frac{n^2}{m(m-1)} I_m=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m$
$\sin n x-I_{m-2}$
or $I_m=\frac{m(m-1)}{n^2}$
$\left[\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x-I_{m-2}\right]$
$I_m=\frac{m}{n^2} x^{m-1} \cos n x+\frac{x^m \sin n x}{n}-\frac{m(m-1)}{n^2} I_{m-2}$
On comparing with Eq. (i), we get
$g(x)=\frac{x^m \sin n x}{n}+\frac{m}{n^2} x^{m-1} \cos n x$
Hence, the correct answer is option: 'D'.