Answer to Question #223112 in Calculus for Kalery

Question

5) ʃ (1-sinx /1-cosx)e^x dx

6) ʃ₀^π/4 cosθ +isinθ / 2cos²θ dθ

Answer 1

5). $\int ({1-sin(x)\over1-cos(x)})e^xdx$

Rewrite to get:

$\int ({sin(x)-1\over cos(x)-1})e^xdx$

Rewriting using trigonometric identities:

$\int ({csc^2({x\over 2})e^x\over 2}-cot({x\over 2})e^x)dx$

Applying linearity we get:

${1\over 2}\int csc^2({x\over 2})e^xdx-\int cot({x\over 2})e^xdx$

Applying integration by parts for $\int cot({x\over 2})e^xdx$

Let $u=cot({x\over 2}) \implies du=-{csc^2({x\over 2})e^x\over 2}dx$

Let $dv=e^xdx \implies v=e^x$

$\int udv=uv-\int vdu$

$=cot({x\over 2})e^x-\int -e^x{csc^2({x\over 2})e^x\over 2}dx$

$=cot({x\over 2})e^x+\int e^x{csc^2({x\over 2})e^x\over 2}dx$

$=cot({x\over 2})e^x+{1\over 2}\int csc^2({x\over 2})e^xdx$

$\implies \int cot({x\over 2})e^xdx=cot({x\over 2})e^x+{1\over 2}\int csc^2({x\over 2})e^xdx$

$\implies {1\over 2}\int csc^2({x\over 2})e^xdx-\int cot({x\over 2})e^xdx={1\over 2}\int csc^2({x\over 2})e^xdx-[cot({x\over 2})e^x+{1\over 2}\int csc^2({x\over 2})e^xdx]$

$={1\over 2}\int csc^2({x\over 2})e^xdx-cot({x\over 2})e^x-{1\over 2}\int csc^2({x\over 2})e^xdx$

The integral ${1\over 2}\int csc^2({x\over 2})e^xdx$ cancels,hence we get

$=-cot({x\over 2})e^x+C$

$\therefore \int ({1-sin(x)\over1-cos(x)})e^xdx=-cot({x\over 2})e^x+C$

6). $\int_0^{\pi\over 4} {cos(\theta)+sin(\theta)\over 2cos^2(\theta)}d\theta$

Rewrite to get:

$=\int_0^{\pi\over 4} ({cos(\theta)\over 2cos^2(\theta)}+{sin(\theta)\over 2cos^2(\theta)})d\theta$

Applying linearity we get:

$={1\over 2}\int_0^{\pi\over 4}{cos(\theta)\over cos^2(\theta)}d\theta+{1\over 2}\int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta$

$={1\over 2}\int_0^{\pi\over 4}{1\over cos(\theta)}d\theta+{1\over 2}\int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta$

$={1\over 2}[\int_0^{\pi\over 4}{1\over cos(\theta)}d\theta+\int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta]$

Applying integration by substitution for $\int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta$ we have

Let $u=cos(\theta) \implies du=-sin(\theta)d\theta$

$\implies \int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta=-\int_0^{\pi\over 4}{du\over u^2}=-\int_0^{\pi\over 4}u^{-2}du=[-u^{-1}]_0^{\pi\over 4}=[-{1\over cos(\theta)}]_0^{\pi\over 4}=-[{1\over cos({\pi\over 4})}-{1\over cos(0)}]={1\over cos(0)}-{1\over cos({\pi\over 4})}$

$\therefore \int_0^{\pi\over 4}{sin(\theta)\over cos^2(\theta)})d\theta={1\over cos(0)}-{1\over cos({\pi\over 4})}=1-1.41421356237309=-0.41421356237309$

Now, let's find $\int_0^{\pi\over 4}{1\over cos(\theta)}d\theta$

${1\over cos(\theta)}=sec(\theta)$

$\implies \int_0^{\pi\over 4}{1\over cos(\theta)}d\theta= \int_0^{\pi\over 4}sec(\theta)d\theta$

Expand fraction by $(tan(\theta)+sec(\theta))$

$\int_0^{\pi\over 4}sec(\theta)d\theta= \int_0^{\pi\over 4}sec(\theta){tan(\theta)+sec(\theta)\over tan(\theta)+sec(\theta)}d\theta$

$= \int_0^{\pi\over 4}{sec(\theta)tan(\theta)+sec^2(\theta)\over tan(\theta)+sec(\theta)}d\theta$

Applying integration by substitution for $\int_0^{\pi\over 4}{sec(\theta)tan(\theta)+sec^2(\theta)\over tan(\theta)+sec(\theta)}d\theta$

Let $u=tan(\theta)+sec(\theta) \implies du=[sec(\theta)tan(\theta)+sec^2(\theta)]d\theta$

$\implies d\theta ={du\over sec(\theta)tan(\theta)+sec^2(\theta)}$

$\implies\int_0^{\pi\over 4}{sec(\theta)tan(\theta)+sec^2(\theta)\over tan(\theta)+sec(\theta)}d\theta=\int_0^{\pi\over 4}{sec(\theta)tan(\theta)+sec^2(\theta)\over u}{du\over sec(\theta)tan(\theta)+sec^2(\theta)}$

The terms $sec(\theta)tan(\theta)+sec^2(\theta)$ cancels hence we get

$=\int_0^{\pi\over 4}{du\over u}=[ln(u)]_0^{\pi\over 4}=[ln(tan(\theta)+sec(\theta))]_0^{\pi\over 4}=[ln(tan({\pi\over 4})+sec({\pi\over 4}))]-[ln(tan(0)+sec(0))]$

$tan({\pi\over 4})=1$

$sec({\pi\over 4})={1\over cos({\pi\over 4})}=1.41421356237309$

$tan(0)=0$

$sec(0)={1\over cos(0)}=1$

$\int_0^{\pi\over 4}{du\over u}=[ln(1+1.41421356237309)]-[ln(0+1)]$

$=[ln(2.41421356237309)]-[ln(1)]=0.88137358701954$

$\implies \int_0^{\pi\over 4} {cos(\theta)+sin(\theta)\over 2cos^2(\theta)}d\theta={1\over 2}[0.88137358701954-0.41421356237309]=0.23358001232323$