Compute ∫x2dx(xsin⁡x+cos⁡x)2

Question

Compute  ∫x2dx(xsin⁡x+cos⁡x)2

Barbara Meeker · Accepted Answer

Observe that, d(xsin⁡x+cos⁡x)dx=xcos⁡x  ∫x2dx(xsin⁡x+cos⁡x)2=∫xcos⁡x⋅xcos⁡x(xsin⁡x+cos⁡x)2dx  So, if z=xsin⁡x+cos⁡x, dz=xcos⁡xdx  So, ∫xcos⁡x(xsin⁡x+sin⁡x)2dx=∫dzz2=−12=−1xsin⁡x+cos⁡x  So,  I=xcos⁡x∫xcos⁡x(xsin⁡x+cos⁡x)2dx−∫(d(xcos⁡x)dx∫xcos⁡x(xsin⁡x+cos⁡x)2dx)dx  =−xcos⁡x(xsin⁡x+cos⁡x)+∫(xsin⁡x+cos⁡xcos2x)(1xsin⁡x+cos⁡x)dx  =−xcos⁡x(xsin⁡x+cos⁡x)+∫sec2xdx  =−xcos⁡x(xsin⁡x+cos⁡x}+tan⁡x+C  where C is an arbitrary constant of indefinite integral  Another forn will be sin

Kayla Kline · Accepted Answer

I was inspired by this post to conduct this method.

\int \frac{x^{2}}{{(x \sin x + \cos x)}^{2}} d x

The Harmonic Addition Theorem comes in handy, so

x \sin x + \cos x = \sqrt{1 + x^{2}} \cos (x - α)

where

α = \arctan (x)

.
The origin integration becomes:

\int \frac{x^{2}}{1 + x^{2}} \sec^{2} (x - α) d x

Notice that

\int \frac{x^{2}}{1 + x^{2}} d x = x - \arctan x = x - α

So let

t = x - α

then

d t = \frac{x^{2}}{1 + x^{2}} d x

and the integration simplifies as:

\int \sec^{2} (t) d t = \tan (t) = \tan (x - \arctan x)

In conclusion,

\int \frac{x^{2}}{{(x \sin x + \cos x)}^{2}} d x = \tan (x - \arctan x) + C

karton · Accepted Answer

Differentiation of xsin⁡x+cos⁡x is cos⁡x ∫xcos⁡x(xsin⁡x+cos⁡x)2⏟II⋅x(cos⁡x)dx Now integrate by parts. I=−1(xsin⁡x+cos⁡x)⋅xcos⁡x+∫1(xsin⁡x+cos⁡x)⋅cos⁡x.1−x(−sin⁡x)cos2⁡x Now, I hope things are clear to you.

Compute \int\frac{x^2dx}{(x\sin x+\cos x)^2}

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