int {{x^{51}}({{tan }^{ - 1}}x + {{cot }^{ - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that for all $x \in \mathbb{R}$,the identity holds: ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$.Substituting this into the integr

Vedclass · Accepted Answer

$\frac{{{x^{52}}}}{{52}}({	an ^{ - 1}}x + {\cot ^{ - 1}}x) + c$

Vedclass · Answer

(A) We know that for all $x \in \mathbb{R}$,the identity holds: ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$.Substituting this into the integral,we get:$\int {{x^{51}}({{	an }^{ - 1}}x + {\cot ^{ - 1}}x)} \,dx = \int {{x^{51}} \cdot \frac{\pi }{2}} \,dx$.Taking the constant $\frac{\pi }{2}$ outside the integral:$= \frac{\pi }{2} \int {{x^{51}}} \,dx$.Using the power rule $\int {{x^n}} \,dx = \frac{{{x^{n + 1}}}}{{n + 1}} + c$:$= \frac{\pi }{2} \cdot \frac{{{x^{52}}}}{{52}} + c = \frac{{\pi {x^{52}}}}{{104}} + c$.Since $\frac{\pi }{2} = {	an ^{ - 1}}x + {\cot ^{ - 1}}x$,we can rewrite the result as:$= \frac{{{x^{52}}}}{{52}} \cdot \frac{\pi }{2} + c = \frac{{{x^{52}}}}{{52}}({	an ^{ - 1}}x + {\cot ^{ - 1}}x) + c$.

$\int {{x^{51}}({{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x)} \,dx = $

Similar Questions

$\int {\frac{{\cos x}}{{\cos (x - a)}}dx - } \int {\frac{{\sin x}}{{\sin (x - a)}}dx = } $

$\int (x + \frac{1}{x})^3 dx = $

$\int {\frac{{1 + x + \sqrt {x + {x^2}} }}{{\sqrt x + \sqrt {1 + x} }}\,dx} = $

Integrate the function $\sin (ax+b) \cos (ax+b)$.

$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module