Evaluate \(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\), by first using the substitution \(u=\dfrac{\pi}{2}-x\). (4 marks)
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\(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\)
\(\text{Let} \ \ u=\dfrac{\pi}{2}-x \ \ \Rightarrow\ \ \dfrac{du}{dx}=-1 \ \ \Rightarrow\ \ du=-dx\)
\(\text{Limits:} \ \ u=\dfrac{\pi}{2}\ \ \Rightarrow\ \ x=0, \ \ u=0\ \ \Rightarrow\ \ x=\dfrac{\pi}{2}\)
| \(I\) | \(=-\displaystyle \large{\int_{\small{\dfrac{\pi}{2}}}^0}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)}\,dx\) |
| \(=\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\cos x+\sin x}\, d x\) |
\(\text{Add \(I\) (swap variable from \(x\) to \(u\)) to original integral:}\)
\(2I=\dfrac{\pi}{2} \displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{1}{1+\sin u+\cos u}\, d u\)
\(\text{Substitute} \ \ t=\tan \left(\frac{u}{2}\right), \ \sin u=\dfrac{2t}{1+t^2}, \ \cos u=\dfrac{1-t^2}{1+t^2}\)
\(d t=\dfrac{1}{2} \sec ^2\left(\frac{u}{2}\right)\, du \ \ \Rightarrow\ \ du=\dfrac{2}{1+\tan ^2\left(\frac{u}{2}\right)}\, dt=\dfrac{2}{1+t^2}\, d t\)
\(\text{Limits:} \ \ n=\dfrac{\pi}{2}\ \Rightarrow \ t=1, \ n=0 \ \Rightarrow \ t=0\)
| \(2I\) | \(=\dfrac{\pi}{2} \displaystyle \int_0^1 \dfrac{1}{1+\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} \times \frac{2}{1+t^2}\,d t\) |
| \(I\) | \(=\displaystyle\frac{\pi}{4} \int_0^1 \frac{2}{1+t^2+2 t+1-t^2}\, d t\) |
| \(=\displaystyle \frac{\pi}{4} \int_0^1 \frac{1}{1+t}\, d t\) | |
| \(=\dfrac{\pi}{4}\Bigl[\ln (1+t)\Bigr]_0^1\) | |
| \(=\dfrac{\pi}{4}(\ln 2-\ln 1)\) | |
| \(=\dfrac{\pi \, \ln 2}{4}\) |