Question

If $I_n=\int ta{n}^{n}xdx$ then which of the following relation is correct -

• A. $I_n=\frac{ta{n}^{n-1}\left(x\right)}{n-1}-I_{n-2}$
• B. $I_n=\frac{ta{n}^{n-2}\left(x\right)}{n-2}-I_{n-2}$
• C. $I_n=\frac{ta{n}^{n-3}\left(x\right)}{n-3}-I_{n-2}$
• D. $I_n=\frac{ta{n}^{n-4}\left(x\right)}{n-4}-I_{n-2}$

Answer 1

The correct option is A $I_n=\frac{ta{n}^{n-1}\left(x\right)}{n-1}-I_{n-2}$

$I_n=\int ta{n}^{n}xdx{I}_n=\int ta{n}^{n-2}x.ta{n}^2\left(x\right)dx$
Or $I_n=\int ta{n}^{n-2}x.\left(se{c}^2\right(x)-1)dx$
Or $I_n=\int ta{n}^{n-2}x.\left(se{c}^2\right(x)dx-\int ta{n}^{n-2}\left(x\right)dx$....(1)
Let $I=\int ta{n}^{n-2}x.\left(se{c}^2\right(x)dx$
Let’s substitute tan(x) = t
$\Rightarrow se{c}^2\left(x\right).dx=dtI=\int t^{n-2}.dt$
Or $I=\frac{t^{n-1}}{n-1}$
Or $I=\frac{ta{n}^{n-1}\left(x\right)}{n-1}$
Substituting $I=\frac{ta{n}^{n-1}\left(x\right)}{n-1}$ in 1st equation.
So, we’ll have
$I_n=\frac{ta{n}^{n-1}\left(x\right)}{n-1}-\int ta{n}^{n-2}\left(x\right)dx$
We can see that the integral $\int ta{n}^{n-2}\left(x\right)dx$ is is nothing but $I_{n-2}$
So, the relation will be
$I_n=\frac{ta{n}^{n-1}\left(x\right)}{n-1}-I_{n-2}$