Question

Find the value of the integration $\int {{e^x}(\sin x + 2\cos x)\sin xdx} $ .
A.${e^x}\cos x + c$
B.${e^x}\sin x + c$
C.${e^x}{\sin ^2}x + c$
D.${e^x}\sin 2x + c$

Answer 1

Hint: Multiply ${e^x}\sin x$to the expression in the braces, then split the integration.
Now, apply integration by parts to the first part and keep the second part as it is and calculate to obtain the required result.

Formula Used:
$\dfrac{d}{{dx}}{(f(x))^n} = n{\left( {f(x)} \right)^{n - 1}}\dfrac{d}{{dx}}f(x)$
$\int {{e^x} = {e^x}} $
$\int {uvdx = } u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\int {vdx} } \right]} } dx$ , where u and v are functions of x only.

Complete step by step solution:
The given integral is,
$\int {{e^x}(\sin x + 2\cos x)\sin xdx} $
Break the above integration into 2 integrations
 $ = \int {\left[ {{e^x}{{\sin }^2}x + 2{e^x}\sin x\cos x} \right]dx} $
$ = \int {{e^x}{{\sin }^2}xdx + \int {2{e^x}\sin x\cos xdx} } $
Apply by parts integration on the first integration and the second integration will be the same as it is
$ = {\sin ^2}x\int {{e^x}dx - \int {[\dfrac{d}{{dx}}({{\sin }^2}x)} } \int {{e^x}dx} ]dx + \int {2{e^x}\sin x\cos xdx} $
$ = {e^x}{\sin ^2}x - \int {2{e^x}\sin x\cos xdx} + \int {2{e^x}\sin x\cos xdx} $
Now cancel the integrations
$ = {e^x}{\sin ^2}x + c$, where c is integrating constant.

Option ‘C’ is correct

Note: Do not get confused with LIATE and ILATE. ILATE is the correct one.
Here use “ILATE” rule to choose ‘u’ for the integration by parts.
I denotes Inverse trigonometry
L denotes Logarithm function
A denotes Algebraic expression
T denotes trigonometry function
E denotes exponential function