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Integration by Parts

Integration by parts: $\int fg' \,dx=fg-\int f'g \,dx.$ Equivalently, $\int u\,dv=uv-\int v \,du.$

By the product rule, $$(fg)'=f'g+fg'$$. Integrating both sides we get, $\int (fg)' \,dx=\int (f'g+fg')\,dx \implies fg=\int f'g \,dx+\int fg'\,dx.$ Subtracting $$\int f'g\,dx$$ from both sides, we get $\int fg' \,dx=fg-\int f'g \,dx.$ The rest follows by the substitution $$u=f$$ and $$dv=g'\,dx$$ for which $$du=f'\,dx$$ and $$v=\int dv=\int g'\,dx=g$$.

Example. \begin{align*} \int xe^x \,dx &= \int u \,dv & (\text{let } u=x,\, dv=e^x \,dx)\\ &= uv-\int v \,du & \left( du=dx,\, v=\int e^x \,dx=e^x \right)\\ &= xe^x-\int e^x \,dx & \\ &= xe^x-e^x+C & \end{align*} Note that the choice of $$u$$ and $$v$$ is important. Integration by parts does not produce any simple answer if we chose $$u=e^x$$ and $$dv=x\,dx$$. Usually we choose $$u$$ so that its derivative is "simpler" than $$u$$ and $$dv$$ is integrable. Some like the following decreasing order of preference for $$u$$: LIATE (Logarithm, Inverse Trig, Algebraic (Polynomial), Trig, Exponential).

Example. \begin{align*} \int \ln x \,dx &= \int u \,dv & (\text{let } u=\ln x,\, dv=dx)\\ &= uv-\int v \,du & \left( du=\frac{1}{x} \,dx,\, v=\int dx=x \right)\\ &= (\ln x)x-\int x\frac{1}{x} \,dx & \\ &= (\ln x)x-\int dx & \\ &= x\ln x-x+C & \end{align*}

Circular integration by parts: Sometimes we need to do integration by parts more than once and get the same integral back.

Example. \begin{align*} \int e^x\sin x \,dx &= \int u \,dv & (\text{let } u=\sin x,\, dv=e^x\,dx)\\ &= uv-\int v \,du & \left( du=\cos x \,dx,\, v=\int e^x \,dx=e^x \right)\\ &= \sin x e^x-\int e^x\cos x \,dx. & (1) \end{align*} Now we apply integration by parts on $$\int e^x\cos x \,dx$$. Let $$u=\cos x,\, dv=e^x\,dx$$. Then $$du=-\sin x \,dx,\, v=\int e^x \,dx=e^x$$. \begin{align*} \int e^x\cos x \,dx &= \int u \,dv & \\ &= uv-\int v \,du & \\ &= \cos x e^x-\int e^x(-\sin x) \,dx & \\ &= \cos x e^x+\int e^x\sin x \,dx. & \end{align*} Plugging this in (1), we get \begin{align*} \int e^x\sin x \,dx &= \sin x e^x-\int e^x\cos x \,dx & \\ &= \sin x e^x-\left( \cos x e^x+\int e^x\sin x \,dx \right) & \\ &= e^x(\sin x -\cos x) -\int e^x\sin x \,dx. & \\ \end{align*} Adding $$\int e^x\sin x \,dx$$ to both sides, we get $\begin{array}{rrcl} & 2\int e^x\sin x \,dx & = & e^x(\sin x -\cos x)+C\\ \implies & \int e^x\sin x \,dx & = & \frac{1}{2}e^x(\sin x -\cos x)+C. \end{array}$

Reduction Formulas: Integrating by parts in a circular fashion we get the following reduction formulas: $\int \cos^n x \,dx=\frac{1}{n} \cos^{n-1}x \sin x+\frac{n-1}{n}\int \cos^{n-2}x \, dx,\, n\geq 2.$ $\int \sin^n x \,dx=-\frac{1}{n} \sin^{n-1}x \cos x+\frac{n-1}{n}\int \sin^{n-2}x \, dx,\, n\geq 2.$

\begin{align*} \int \cos^n x \,dx &= \int u \,dv \,\, \left(\text{let } u=\cos^{n-1}x,\, dv=\cos x \,dx \right)\\ &= uv-\int v \,du \,\, \left( du=(n-1) \cos^{n-2} x(-\sin x)\,dx,\, v=\int \cos x \,dx=\sin x \right)\\ &= \cos^{n-1}x \sin x-\int \sin x (n-1) \cos^{n-2} x (-\sin x)\,dx \\ &= \cos^{n-1}x \sin x+(n-1) \int \sin^2 x \cos^{n-2} x \,dx \\ &= \cos^{n-1}x \sin x+(n-1) \int (1-\cos^2 x) \cos^{n-2} x \,dx \\ &= \cos^{n-1}x \sin x+(n-1) \int (\cos^{n-2} x- \cos^n x) \,dx \\ &= \cos^{n-1}x \sin x+(n-1) \int \cos^{n-2} x \,dx- (n-1) \int \cos^n x \,dx \end{align*} Adding $$(n-1) \int \cos^n x \,dx$$ to both sides, we get \begin{align*} & n\int \cos^n x \,dx &&= \cos^{n-1}x \sin x+(n-1) \int \cos^{n-2} x \,dx\\ \implies & \int \cos^n x \,dx &&=\frac{1}{n} \cos^{n-1}x \sin x+\frac{n-1}{n}\int \cos^{n-2}x \, dx. \end{align*} The other formula has a similar proof.

Example. \begin{align*} \int \cos^4 x \,dx &= \frac{1}{4} \cos^{3}x \sin x+\frac{3}{4}\int \cos^{2}x \, dx &\\ \end{align*} We can also apply the reduction formula to $$\int \cos^2 x \,dx$$. Alternatively we can use the trig identity $$\cos^2 x=\displaystyle\frac{1+\cos(2x)}{2}$$. Then \begin{align*} \int \cos^4 x \,dx &= \frac{1}{4} \cos^{3}x \sin x+\frac{3}{4}\int \cos^{2}x \, dx &\\ &= \frac{1}{4} \cos^{3}x \sin x+\frac{3}{8}\int (1+\cos(2x)) \, dx &\\ &= \frac{1}{4} \cos^{3}x \sin x+\frac{3}{8} \left( x+\frac{\sin(2x)}{2} \right) +C &\\ \end{align*}

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