SATHEE: Definite Integration Question 66

Definite Integration Question 66

Question 66

Evaluate

$\int_{0}^{π} e^{| \cos x |} 2 \sin \frac{1}{2} \cos x + 3 \cos \frac{1}{2} \cos x \sin x d x$ .

$(2005, 2 M)$

Show Answer

Solution:

Let

$I = \int_{0}^{π} e^{| \cos x |} 2 \sin \frac{1}{2} \cos x + 3 \cos \frac{1}{2} \cos x \sin x d x$

$\Rightarrow I = \int_{0}^{π} e^{| \cos x |} \cdot \sin x \cdot 2 \sin \frac{1}{2} \cos x d x$

$+ \int_{0}^{π} e^{| \cos x |} \cdot 3 \cos \frac{1}{2} \cos x \cdot \sin x d x$

$\Rightarrow I = I_{1} + I_{2}$ using $\int_{0}^{2 a} f (x) d x$

$0, f (2 a - x) = - f (x)$

$= 2 \int_{0}^{a} f (x) d x, f (2 a - x) = + f (x)$

where, $I_{1} = 0 [∵ f (π - x) = - f (x)]$

and $I_{2} = 6 \int_{0}^{π / 2} e^{\cos x} \cdot \sin x \cdot \cos \frac{1}{2} \cos x d x$

Now, $I_{2} = 6 \int_{0}^{1} e^{t} \cdot \cos \frac{t}{2} d t$

[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

$I = \frac{24}{5} e \cos \frac{1}{2} + \frac{e}{2} \sin \frac{1}{2} - 1$

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