\hspace{-16}(1)\;\;\int\frac{\sin x+\cos x}{\sin^2 x+\cos ^4 x}dx\\\\\\ (2)\;\; \int\frac{\cos x+x.\sin x}{x^2+\cos^2 x}dx
asked in goiit
341I beg to differ
\sin^2 x + \cos^4 x = 1 - \sin^2 x \cos^2 x
Now
\sin^2 x \cos^2 x = \left(\frac{t^2-1}{2} \right)^2
where t = \sin x - \cos x
Hence the given integral becomes
\int \frac{1}{1-\left(\frac{t^2-1}{2} \right)^2 } \ dt = \int \frac{1}{\left(\frac{3}{2}-t^2\right)\left(\frac{1}{2}+t^2\right)} \ dt
This is easily handled
1708Thanks sir.(very easy.)
for (1)
http://www.goiit.com/posts/list/integral-calculus-integration-of-sinx-cosx-sin-2x-cos-4x-1119365.htm
for (2)
http://www.goiit.com/posts/list/integral-calculus-please-give-me-the-answer-of-this-integral-1118958.htm#1529259
1357Second one
Multiply Nr and Dr by sec2x and then put x secx =t
