Question

evaluate the following $\int\limits^1_0 {\frac{sin(3 tan^-1 x)tan^-1 x}{1+x^2} } \, dx$


if you give silly answer I will report you

Answer 1

Step-by-step explanation:

We have,

$\int \limits^{1}_{0} \frac{ \sin(3 \tan^{ - 1} (x) ) \tan^{ - 1} (x) }{1 + {x}^{2} } dx$

$let \: \: \tan^{ - 1} (x) = t \\ \implies \frac{dx}{1 + {x}^{2} } = dt$

$\int \limits^{ \frac{\pi}{4} } _{0} { \sin(3t). t}dt$

Firstly, solving the integral by integration by parts and then useing the limits, we have,

$[(\int \sin(3t) dt)\times t]^{ \frac{\pi}{4} }_{0} - \int \limits^{ \frac{\pi}{4} } _{0}( \frac{d}{dt}(t) (\int( \sin(3t) )dt)dt )$

$\implies [ \frac{ \cos(3t)}{3} \times t]^{ \frac{\pi}{4} }_{0} - \int \limits^{ \frac{\pi}{4} } _{0}1 \times - \frac{ \cos(3t) }{3} dt$

$\implies( \frac{1}{3} \cos( \frac{3\pi}{4} ) \times \frac{\pi}{4} - 0) + \frac{1}{3} \int \limits^{ \frac{\pi}{4} } _{0} \cos(3t) dt$

$\implies - \frac{\pi}{12 \sqrt{2} } + \frac{1}{3} [ \frac{ \sin(3t) }{3} ]^{ \frac{\pi}{4} } _{0}$

$\implies - \frac{\pi}{12 \sqrt{2} } + \frac{1}{9} ( \sin( \frac{3\pi}{4} ) - 0)$

$\implies - \frac{\pi}{12 \sqrt{2} } + \frac{1}{9 \sqrt{2} }$