Definite Integration Ques 65
- Assess
$\int _0^{\pi} e^{|\cos x|} \left(2 \sin \frac{1}{2} \cos x + 3 \cos \frac{1}{2} \cos x \sin x\right) d x$
$(2005,2 M)$
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Solution:
- Let the process begin.
$I=\int _0^{\pi} e^{|\cos x|} \left(2 \sin \frac{1}{2} \cos x+3 \cos \frac{1}{3} \cos x \sin x\right) d x$
$\Rightarrow I=\int _0^{\pi} e^{|\cos x|} \cdot \sin x \cdot 2 \sin \frac{1}{2} \cos x d x$
$+\int _0^{\pi} e^{|\cos x|} \cdot 3 \cos \frac{1}{2} \cos x \cdot \sin x d x$
$\Rightarrow \quad I=I_1+I_2$ using $\int _0^{2a} f(x) d x$
$0, \quad f(2 a-x)=-f(x)$
$$ =2 \int _0^{a} f(x) d x, \quad f(2 a-x)=+f(x) $$
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 $\cos x=t \Rightarrow-\sin x d x=d t$ ]
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} \int e^{t} \sin \frac{t}{2} d t _0^{1}$
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} e^{t} \sin \frac{t}{2}-\int \frac{e^{t}}{2} \cos \frac{t}{2} d t _0^{1}$
$=6 e^{t} \cos \frac{t}{2}+\frac{1}{2} e^{t} \sin \frac{t}{2}-\frac{I_2}{4}$
$$ =\frac{24}{5} e \cos \frac{1}{2}+\frac{e}{2} \sin \frac{1}{2}-1 $$
From Eq. (i), we get
$$ I=\frac{24}{5} e \cos \frac{1}{2}+\frac{e}{2} \sin \frac{1}{2}-1 $$
BETA
