The correct option is B ^n+1xn+1+c I = ∫(^n+2x+^nx)dx = ∫^nx.cosec ^2xdx Put #160; z = x ⇒ #160;dz = cosec ^2 x dx ∴ I = ∫ z^ndz = ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Substitution Method to Remove Indeterminate Form

∫n+2x+nxdx=

Question

$\int ({cot}^{n + 2} x + {cot}^{n} x) d x =$

\frac{- {cot}^{n + 1} x}{n + 1} + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{- {cot}^{n - 1} x}{n - 1} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- {cot}^{n + 3} x}{n + 3} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- {cot}^{2 n} x}{2 n} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{- {cot}^{n + 1} x}{n + 1} + c$
$I = \int ({cot}^{n + 2} x + {cot}^{n} x) d x = \int {cot}^{n} x . {cosec}^{2} x d x$
Put $z = cot x \Rightarrow d z = - {cosec}^{2} x d x$
$∴ I = \int - z^{n} d z = - \frac{z^{n + 1}}{n + 1} + c = - \frac{{cot}^{n + 1} x}{n + 1} + c$

Suggest Corrections

Similar questions

The (n+1)th term from the end in the expansion of $(2 x - \frac{1}{x})^{3 n}$ is:

$\int \frac{c o s^{n - 1} x}{s i n^{n + 1} x} d x, n \neq 0$ is

\frac{1}{x}

{(1 + x)}^{n} {(1 + \frac{1}{x})}^{n}

\frac{n!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{(2 n - 1)! (2 n + 1)!}

n \geq r - 1

{^{n} C}_{r - 1} + {^{n + 1} C}_{r - 1} + {^{n + 2} C}_{r - 1} + . . . . + {^{2 n} C}_{r - 1} = {^{2 n + 1} C}_{r^{2} - 182} - {^{n} C}_{r}

n

n

^{2 n} C_{3} :^{n} C_{3} = 12 : 1

Join BYJU'S Learning Program