Question

Integration of tan^-1(1+cosx/sinx)dx

Answer 1

hope this answer is helpful plz mark it as brainliest if u think I can help u just follow me

Answer 2

$\boxed{\underline{\textsf{Formulas :}}}$

$1.\:\bold{sin2x = 2sinx cosx}$

$2.\:\bold{cos2x=cos^{2}x-sin^{2}x}$

$3.\:\bold{cotx=\frac{cosx}{sinx}}$

$4.\:\bold{cotx=tan(\frac{\pi}{2}-x)}$

$5.\: \bold{\int x^{n}=\frac{x^{n+1}}{n+1}+C}$

$\textsf{where C is integral constant.}$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\: \bold{\dfrac{1+cosx}{sinx}}$

$=\bold{\small{\dfrac{sin^{2}\frac{x}{2}+cos^{2}\dfrac{x}{2}+cos^{2}\dfrac{x}{2}-sin^{2}\dfrac{x}{2}}{2sin\dfrac{x}{2}cos\dfrac{x}{2}}}}$

$=\bold{\dfrac{2cos^{2}\dfrac{x}{2}}{2sin\dfrac{x}{2}cos\dfrac{x}{2}}}$

$=\bold{\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}}$

$=\bold{cot\dfrac{x}{2}}$

$=\bold{tan(\dfrac{\pi}{2}-\dfrac{x}{2})}$

$\to \bold{\dfrac{1+cosx}{sinx} = tan(\dfrac{\pi}{2}-\dfrac{x}{2})}$

$\to \bold{\bold{\tiny{tan^{-1}(\dfrac{1+cosx}{sinx})= tan^{-1}\{tan(\dfrac{\pi}{2}-\dfrac{x}{2})\}}}}$

$\to \bold{tan^{-1}\{\dfrac{1+cosx}{sinx}\}=\dfrac{\pi}{2}-\dfrac{x}{2}}$

$\textsf{On integration, we get}$

$\int \bold{tan^{-1}(\frac{1+cosx}{sinx})dx=\int (\frac{\pi}{2}-\frac{x}{2})dx}$

$=\bold{\frac{\pi}{2} \int dx - \frac{1}{2} \int xdx}$

$=\bold{\frac{\pi}{2}x - \frac{1}{2}*\frac{x^{2}}{2}+C}\:,$

$\textsf{where C is integral constant}$

$=\bold{\frac{\pi}{2}x - \frac{x^{2}}{4}+C}$

$\to \boxed{\boxed{\bold{\tiny{\int tan^{-1}(\frac{1+cosx}{sinx})dx = \frac{\pi}{2}x - \frac{x^{2}}{4}+C}}}}$

$\textsf{which is the required integral.}$