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)
\(I=\frac{{{e}^{t}}}{2}\left( \sin t-\cos t \right)+C\),
\(\int \sin \left( \log x \right)dx=\frac{x}{2}\left[ \sin \left( \log x \right)-\cos \left( \log x \right) \right]+C\).