Frankenstein?

Calculus Level 4

∫ 0 π e x sin ⁡ 2 ( x ) d x \large \int_0^\pi e^x \sin^2(x) \, dx

The above integral has a simple exact closed form. Evaluate this integral and give your answer to 3 decimal places.


The answer is 8.856.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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3 solutions

Chew-Seong Cheong
Feb 13, 2016

∫ 0 π e x sin ⁡ 2 x d x = 1 2 ∫ 0 π e x ( 1 − cos ⁡ ( 2 x ) ) d x = 1 2 ∫ 0 π e x d x − 1 4 ∫ 0 π e x ( e 2 x i + e − 2 x i ) d x = 1 2 [ e x ] 0 π − 1 4 [ e ( 1 + 2 i ) x 1 + 2 i + e ( 1 − 2 i ) x 1 − 2 i ] 0 π = 1 2 ( e π − 1 ) − 1 4 [ e x ( e 2 x i + e − 2 x i − 2 i e 2 x i + 2 i e − 2 x i ) 1 + 4 ] 0 π = 1 2 ( e π − 1 ) − 1 2 [ e x ( cos ⁡ ( 2 x ) + 2 i sin ⁡ ( 2 x ) ) 5 ] 0 π = 1 2 ( e π − 1 ) − 1 10 [ e π ( cos ⁡ ( 2 π ) + 2 i sin ⁡ ( 2 π ) ) − e 0 ( cos ⁡ 0 + 2 i sin ⁡ 0 ) ] = 1 2 ( e π − 1 ) − 1 10 ( e π − 1 ) = 2 5 ( e π − 1 ) = 8.856 \begin{aligned} \int_0^\pi e^x \sin^2 x \space dx & = \frac{1}{2} \int_0^\pi e^x (1-\cos(2x)) \space dx \\ & = \frac{1}{2} \int_0^\pi e^x \space dx - \frac{1}{4} \int_0^\pi e^x \left(e^{2xi} + e^{-2xi} \right) \space dx \\ & = \frac{1}{2}\left[e^x\right]_0^\pi - \frac{1}{4}\left[\frac{e^{(1+2i)x}}{1+2i} + \frac{e^{(1-2i)x}}{1-2i}\right]_0^\pi \\ & = \frac{1}{2}\left(e^\pi -1\right) - \frac{1}{4}\left[\frac{e^x\left(e^{2xi} + e^{-2xi} -2ie^{2xi} + 2i e^{-2xi} \right)}{1+4} \right]_0^\pi \\ & = \frac{1}{2}\left(e^\pi -1\right) - \frac{1}{2}\left[\frac{e^x\left(\cos (2x) + 2i\sin (2x) \right)}{5} \right]_0^\pi \\ & = \frac{1}{2}\left(e^\pi -1\right) - \frac{1}{10}\left[e^\pi \left(\cos (2\pi) + 2i\sin (2\pi) \right) - e^0 \left(\cos 0 + 2i\sin 0 \right) \right] \\ & = \frac{1}{2}\left(e^\pi -1\right) - \frac{1}{10}\left(e^\pi - 1 \right) \\ & = \frac{2}{5}\left(e^\pi -1\right) = \boxed{8.856} \end{aligned}

10 Helpful 0 Interesting 2 Brilliant 0 Confused

certainly easy solution took me long to solve this, how fast do you solve the problem and how do you realize to use certain things instead others?

Mardokay Mosazghi - 5 years, 4 months ago

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I like to use e i θ = cos ⁡ θ + i sin ⁡ θ e^{i\theta} = \cos \theta + i \sin \theta . Frequently, it simplifies matter. In this case integration of e z e^{z} is simple.

Chew-Seong Cheong - 5 years, 4 months ago

Exponential + trig combination functions always scream to me to use Euler's Cosine & Sine Identities as a simple substitution (much like Chew-Seong's hint above). That helps transform the above integrand into a series of exponential function terms that is readily integrable.

tom engelsman - 5 years, 4 months ago

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yes sir i am also learning to be aware when yo use Euler cosines and sin identities.

Mardokay Mosazghi - 5 years, 4 months ago

Yes, Euler is my greatest mathematician.

Chew-Seong Cheong - 5 years, 4 months ago

Great solution !

Keshav Tiwari - 5 years, 4 months ago
Josh Banister
Feb 13, 2016

To do this problem the way I did, we first find the antiderivative of e x sin ⁡ 2 ( x ) e^x \sin^2(x) and then using the fundamental theorem of calculus, solve this problem. Firstly though, we use integration by parts a few times ∫ e x sin ⁡ 2 ( x ) d x = e x sin ⁡ 2 ( x ) − ( ∫ e x ( 2 sin ⁡ ( x ) cos ⁡ ( x ) ) d x ) = e x sin ⁡ 2 ( x ) − 2 ∫ e x sin ⁡ ( x ) cos ⁡ ( x ) d x = e x sin ⁡ 2 ( x ) − 2 ( e x sin ⁡ ( x ) cos ⁡ ( x ) − ∫ e x ( cos ⁡ 2 ( x ) − sin ⁡ 2 ( x ) ) d x ) = e x sin ⁡ 2 ( x ) − 2 e x sin ⁡ ( x ) cos ⁡ ( x ) + 2 ∫ e x ( 1 − 2 sin ⁡ 2 ( x ) ) d x = e x sin ⁡ 2 ( x ) − 2 e x sin ⁡ ( x ) cos ⁡ ( x ) + 2 e x − 4 ∫ e x sin ⁡ 2 ( x ) d x ⟹ 5 ∫ e x sin ⁡ 2 ( x ) d x = e x sin ⁡ 2 ( x ) − 2 e x sin ⁡ ( x ) cos ⁡ ( x ) + 2 e x ⟹ ∫ e x sin ⁡ 2 ( x ) d x = 1 5 ( e x sin ⁡ 2 ( x ) − 2 e x sin ⁡ ( x ) cos ⁡ ( x ) + 2 e x ) \begin{aligned} \int e^x \sin^2(x) dx &= e^x \sin^2(x) - \left( \int e^x \big(2 \sin(x) \cos(x)\big) dx \right) \\ &= e^x \sin^2(x) - 2\int e^x \sin(x) \cos(x) dx \\ &= e^x \sin^2(x) - 2\left( e^x \sin(x) \cos(x) - \int e^x \big(\cos^2(x) - \sin^2(x) \big) dx \right) \\ &= e^x \sin^2(x) - 2e^x \sin(x) \cos(x) + 2\int e^x \big(1 - 2\sin^2(x) \big) dx \\ &= e^x \sin^2(x) - 2e^x \sin(x) \cos(x) + 2e^x - 4\int e^x \sin^2(x) dx \\ \implies 5 \int e^x \sin^2(x) dx &= e^x \sin^2(x) - 2e^x \sin(x) \cos(x) + 2e^x \\ \implies \int e^x \sin^2(x) dx &= \frac{1}{5} \left(e^x \sin^2(x) - 2e^x \sin(x) \cos(x) + 2e^x \right) \end{aligned}

Let F ( x ) = 1 5 ( e x sin ⁡ 2 ( x ) − 2 e x sin ⁡ ( x ) cos ⁡ ( x ) + 2 e x ) F(x) = \frac{1}{5} \left(e^x \sin^2(x) - 2e^x \sin(x) \cos(x) + 2e^x \right) . By direct calculation (and realising that sin ⁡ ( 0 ) = sin ⁡ ( π ) = 0 \sin(0) = \sin(\pi) = 0 ), we have F ( 0 ) = 2 5 F(0) = \frac{2}{5} and F ( π ) = 2 5 e π F(\pi) = \frac{2}{5}e^{\pi} . And hence we have our final answer. ∫ 0 π e x sin ⁡ 2 ( x ) d x = F ( π ) − F ( 0 ) = 2 5 ( e π − 1 ) \int_0^\pi e^x \sin^2(x) dx = F(\pi) - F(0) = \frac{2}{5} \left( e^\pi - 1\right) Which rounds to 8.856 \boxed{8.856}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

i did the same old boring thing

Mardokay Mosazghi - 5 years, 4 months ago
Kevin Sullivan
Feb 12, 2016

I got the answer, but it took like two pages of work with trig identities and IBP, was there something I missed?

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Try to solve it using complex numbers. Do refer to Chew-Seong Cheong's solution.

Pulkit Gupta - 5 years, 4 months ago

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