Integration by parts | Solution with procedure

9 Oct 2019 Leave a Comment

Here you will see how to solve these integrals https://atomic-temporary-167386734.wpcomstaging.com/integration-by-parts-exercises/ and check if you got them right!

$\int x\sin(x)dx$ $u=x\; \; \; \; \; \; dv=\sin(x)dx$ $du=dx\; \; \; \; \; \; v=-\cos(x)$

$\int x\sin(x)dx=-x\cos(x)+\sin(x)+C$ $\int x\ln(x)dx$ $u=\ln(x)\;\;\;\;\; dv=xdx$ $du=\frac{1}{x}dx\;\;\;\;\; v=\frac{1}{2}x^2$ $\frac{1}{2}x^2\ln\left | x \right |-\frac{1}{2}\int xdx=\frac{1}{2}x^2\ln\left | x \right |-\frac{1}{4}x^2+C=\frac{1}{2}x^2\left ( \ln\left | x \right |-\frac{1}{2} \right )+C$

$\int xe^xdx$ $u=x\;\;\;\;\; dv=e^xdx$ $du=dx\;\;\;\;\; v=e^x$ $xe^x-\int e^xdx=xe^x-e^x=e^x(x-1)+C$ $\int \frac{1}{2}x^2e^xdx=\frac{1}{2}\int x^2e^xdx$ $u=x^2\;\;\;\;\; dv=e^xdx$ $du=2x\,dx\;\;\;\;\; v=e^x$ $\frac{1}{2}x^2e^x-\frac{1}{2}\int2xe^xdx=\frac{1}{2}x^2e^x-\int xe^xdx$ $\frac{1}{2}x^2e^x-\int xe^xdx=\frac{1}{2}x^2e^x-e^x(x-1)=e^x\left ( \frac{1}{2}x^2-x+1\right )+C$

$\int \ln(x)dx$ $u=\ln(x)\;\;\;\;dv=dx$ $du=\frac{1}{x}dx\;\;\;\;v=x$ $x\ln(x)-\int dx=x\ln(x)-x=x(\ln\left | x \right |-1)+C$

$\int \ln^2(x)dx=\int \ln(x)\ln(x)dx$ $u=\ln(x)\;\;\;\;\;dv=\ln(x)dx$ $du=\frac{1}{x}dx\;\;\;\;\;v=x(\ln(x)-1)$ $x(\ln(x)-1)\ln(x)-\int (\ln(x)-1)dx=x\ln^2(x)-x\ln(x)+x-\int \ln(x)dx=$ $x\ln^2(x)-x\ln(x)+x-x(\ln(x)-1)=x(\ln^2(x)-2\ln(x)+2)+C$

$\int \sin(x)\cos(x)dx$ $u=\sin(x)\;\;\;\;\;dv=\cos(x)dx$ $du=\cos(x)\;\;\;\;\;v=\sin(x)$ $\sin^2(x)-\int \sin(x)\cos(x)dx$

Wait… we have the integral of sin(x)cos(x) again! So this means that

$\int \sin(x)\cos(x)dx=\sin^2(x)-\int \sin(x)\cos(x)dx$

and

$2\int \sin(x)\cos(x)dx=\sin^2(x)\rightarrow \int \sin(x)\cos(x)dx=\frac{1}{2}\sin^2(x)+C$

Easy, right?

$\int e^x\cos(x)dx$ $u=e^x\;\;\;\;\;dv=\cos(x)dx$ $du=e^xdx\;\;\;\;\;v=\sin(x)$ $e^x\sin(x)-\int e^x\sin(x)dx$ $\int e^x\sin(x)dx$

Let’s solve this integral by performing another integration by parts.

$u=e^x\;\;\;\;\;dv=\sin(x)dx$ $du=e^xdx\;\;\;\;\;v=-\cos(x)$ $\int e^x\sin(x)dx=-e^x\cos(x)+\int e^x\cos(x)dx$

so we have

$\int e^x\cos(x)dx=e^x\sin(x)+e^x\cos(x)-\int e^x\cos(x)dx$

We can do the same thing we did for the integral of sin(x)cos(x).

$2\int e^x\cos(x)dx=e^x\sin(x)+e^x\cos(x)=e^x(\sin(x)+\cos(x))$

Therefore,

$\int e^x\cos(x)dx=\frac{1}{2}e^x(\sin(x)+\cos(x))+C$

$\int e^x\sin(x)dx$

We now know that this is equal to

$-e^x\cos(x)+\int e^x\cos(x)dx$

which is equal to

$-e^x\cos(x)+\frac{1}{2}e^x(\sin(x)+\cos(x))=e^x\left ( -\cos(x)+\frac{1}{2}\sin(x)+ \frac{1}{2}\cos(x)\right )=$ $\frac{1}{2}e^x(\sin(x)-\cos(x))$

The answer is

$\int e^x\sin(x)dx=\frac{1}{2}e^x(\sin(x)-\cos(x))+C$

$\int \frac{\ln(x)}{x}dx=\int \ln(x)\frac{1}{x}dx$ $u=\ln(x)\;\;\;\;\;dv=\frac{1}{x}dx$ $du=\frac{1}{x}dx\;\;\;\;\;v=\ln(x)$ $\ln^2(x)-\int \frac{\ln(x)}{x}dx$

Same thing here;

$2\int \frac{\ln(x)}{x}dx=\ln^2(x)$ $\int \frac{\ln(x)}{x}dx=\frac{1}{2}\ln^2(x)+C$

And that’s it! Stay tuned for more integrals. You might find these links useful:

https://addjustabitofpi.wordpress.com/2019/10/03/fundamental-integrals/ https://addjustabitofpi.wordpress.com/2019/10/03/rules-and-properties-of-integrals/ https://addjustabitofpi.wordpress.com/2019/10/05/integration-techniques-integration-by-parts/.

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Integration, Integration techniques, Solutions Calculus, Indefinite integrals, integration by parts, integration by parts solutions with procedure

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Integration by parts | Solution with procedure

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Integration by parts | Solution with procedure

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