Definite Integration Question 66
Question 66
- Evaluate
$\int_{0}^{\pi} e^{|\cos x|} 2 \sin \frac{1}{2} \cos x+3 \cos \frac{1}{2} \cos x \sin x d x$.
$(2005,2 \mathrm{M})$
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Solution:
- Let
$I=\int_{0}^{\pi} e^{|\cos x|} 2 \sin \frac{1}{2} \cos x+3 \cos \frac{1}{2} \cos x \sin x d x$
$\Rightarrow I=\int_{0}^{\pi} e^{|\cos x|} \cdot \sin x \cdot 2 \sin \frac{1}{2} \cos x d x$
$+\int_{0}^{\pi} e^{|\cos x|} \cdot 3 \cos \frac{1}{2} \cos x \cdot \sin x d x$
$\Rightarrow \quad I=I_{1}+I_{2}$ using $\int_{0}^{2 a} f(x) d x$
$0, \quad f(2 a-x)=-f(x)$
$$ =2 \int_{0}^{a} f(x) d x, \quad f(2 a-x)=+f(x) $$
where, $\quad I_{1}=0 \quad[\because f(\pi-x)=-f(x)]$
and $\quad I_{2}=6 \int_{0}^{\pi / 2} e^{\cos x} \cdot \sin x \cdot \cos \frac{1}{2} \cos x d x$
Now, $\quad I_{2}=6 \int_{0}^{1} e^{t} \cdot \cos \frac{t}{2} d t$
[put $\cos x=t \Rightarrow-\sin x d x=d t$ ]
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} \int e^{t} \sin \frac{t}{2} d t_{0}^{1}$
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} e^{t} \sin \frac{t}{2}-\int \frac{e^{t}}{2} \cos \frac{t}{2} d t_{0}^{1}$
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} e^{t} \sin \frac{t}{2}^{1}-\frac{I_{2}}{4}$
$$ =\frac{24}{5} e \cos \frac{1}{2}+\frac{e}{2} \sin \frac{1}{2}-1 $$
From Eq. (i), we get
$$ I=\frac{24}{5} e \cos \frac{1}{2}+\frac{e}{2} \sin \frac{1}{2}-1 $$
