# Integration by Parts – Problems

## Integration by Parts – Problems

Integrals of the form ∫eax sin bx dx and ∫eax cos bx dx

In order to evaluate this type of integrals, we use integration by parts as illustrated below.

I = ∫eax sin bx dx

$$=-{{e}^{ax}}\frac{\cos bx}{b}-\int a{{e}^{ax}}\left( \frac{-\cos bx}{b} \right)dx$$,

$$=-\frac{1}{b}{{e}^{ax}}\cos bx+\frac{a}{b}\int {{e}^{ax}}\cos bxdx$$,

$$=-\frac{1}{b}{{e}^{ax}}\cos bx+\frac{a}{b}\left[ {{e}^{ax}}\frac{\sin bx}{b}-\int a{{e}^{ax}}\frac{\sin xbx}{b}dx \right]$$,

$$=-\frac{1}{b}{{e}^{ax}}\cos bx+\frac{a}{{{b}^{2}}}{{e}^{ax}}\sin bx-\frac{{{a}^{2}}}{{{b}^{2}}}\int {{e}^{ax}}\sin bxdx$$,

$$=-\frac{1}{b}{{e}^{ax}}\cos bx+\frac{a}{{{b}^{2}}}{{e}^{ax}}\sin bx-\frac{{{a}^{2}}}{{{b}^{2}}}I$$,

∴ $$I+I.\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{e}^{ax}}}{{{b}^{2}}}\left( a\sin bx-b\cos bx \right)$$,

$$I\left( \frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \right)=\frac{{{e}^{ax}}}{{{b}^{2}}}\left( a\sin bx-b\cos bx \right)$$,

$$I=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin bx-b\cos bx \right)+C$$,

Thus

$$\int {{e}^{ax}}\sin bxdx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin bx-b\cos bx \right)+C$$,

Similarly,

$$\int {{e}^{ax}}\cos bxdx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\cos bx+b\sin bx \right)+C$$.

Example: ∫sin (log x) dx

Solution: ∫sin (log x) dx

where t = log x → x = et

differentiation with respect to x

dt = 1/x dx

Let I = ∫sin (log x) dx then,

I = ∫sin tet dt

I = – et cos t – ∫et (- cos t) dt

I = – et cos t + ∫et cos t dt

I = – et cos t + [et sint – ∫et sin t dt]

I = – et cos t + et sint – I

2I = et (sint – cos t)

$$I=\frac{{{e}^{t}}}{2}\left( \sin t-\cos t \right)+C$$,

$$\int \sin \left( \log x \right)dx=\frac{x}{2}\left[ \sin \left( \log x \right)-\cos \left( \log x \right) \right]+C$$.