Evaluate $\int_{0}^{\pi} e^{2 x} \cdot \sin \left(\frac{\pi}{4}+x\right) d x$

$-\frac{e^{2 \pi}+1}{5 \sqrt{2}}$

CBSE Class 12 · 2016 · Definite IntegralsLong Answer

Evaluate

\int_{0}^{π} e^{2 x} \cdot sin (\frac{π}{4} + x) d x

Answer: $- \frac{e ^{2 π} + 1}{5 2}$

Use the standard result

\int e^{a x} sin (b x + c) d x = \frac{e ^{a x}}{a ^{2} + b ^{2}} [a sin (b x + c) - b cos (b x + c)]

with

a = 2, b = 1, c = \frac{π}{4}

, so

a^{2} + b^{2} = 5

I = \frac{1}{5} [e^{2 x} (2 sin (x + \frac{π}{4}) - cos (x + \frac{π}{4}))]_{0}^{π}

x = π

sin \frac{5 π}{4} = cos \frac{5 π}{4} = - \frac{1}{2}

, so the bracket is

2 (- \frac{1}{2}) - (- \frac{1}{2}) = - \frac{1}{2}

, giving

- \frac{e ^{2 π}}{2}

.
At

x = 0

2 \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{2}

I = \frac{1}{5} (- \frac{e ^{2 π}}{2} - \frac{1}{2}) = - \frac{e ^{2 π} + 1}{5 2}

CBSE Class 12 Board · 2016 · Subjective