_ ^ e^ - x cose c ^2 (2 e^ - x + 5) ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^{ - x}}{\rm{cose}}{{\rm{c}}^2}(2{e^{ - x}} + 5)} \;dx = $

✓
$\frac{1}{2}\cot (2{e^{ - x}} + 5) + c$
B
$ - \frac{1}{2}\cot (2{e^{ - x}} + 5) + c$
C
$2\cot (2{e^{ - x}} + 5) + c$
D
$ - 2\cot (2{e^{ - x}} + 5) + c$

✓

Answer

Correct option: A.

$\frac{1}{2}\cot (2{e^{ - x}} + 5) + c$

a
(a) Put $2{e^{ - x}} + 5 = t \Rightarrow - 2{e^{ - x}}dx = dt,$ then
$\int_{}^{} {{e^{ - x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}(2{e^{ - x}} + 5)\,dx} = - \frac{1}{2}\int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}t\,dt} $
$ = \frac{1}{2}\cot t = \frac{1}{2}\cot (2{e^{ - x}} + 5) + c$.

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