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Integrate [sinx] / (2+sin 2x).

\(\begin{array}{l}I=\int \frac{sinx}{2+sin2x}dx\end{array} \) ⇒ \(\begin{array}{l}\frac{1}{2}\int \frac{sinx+cosx+sinx-cosx}{2+sin2x}dx\end{array} \) ⇒ \(\begin{array}{l}\frac{1}{2}\left [\frac{sinx+cosx}{3-[sinx-cosx]^2} \right ]+\frac{1}{2}\left [\frac{sinx+cosx}{1+[sinx+cosx]^2} \right ]\end{array} \) Now put sinx – cosx =... View Article

Integrate [x-sin x] / [1- cos x].

The given equation is \(\begin{array}{l} \int \frac{x-\sin x}{1-\cos x} dx[\latex] Solution \(\begin{array}{l} \int \frac{x-\sin x}{1-\cos x}[\latex] On using double angle formula for sin... View Article

Find the integral of xlogx/(1+x2)2 under the limits 0 to infinity.

Answer: \(\begin{array}{l}\int \frac{x\ln \left(x\right)}{\left(1+x^2\right)^2}dx\end{array} \). \(\begin{array}{l}\frac{x\ln \left(x\right)}{\left(1+x^2\right)^2}=\left(x\ln \left(x\right)\right)\frac{1}{\left(1+x^2\right)^2}\end{array} \). \(\begin{array}{l}=\int x\ln \left(x\right)\frac{1}{\left(1+x^2\right)^2}dx\end{array} \). Apply Integration By Parts \(\begin{array}{l}u=\ln \left(x\right),:v’=\frac{1}{\left(1+x^2\right)^2}x\end{array} \).... View Article