Question

$\int e^{7x}\sin 5x=\frac{e^{7x}}{\sqrt{74}}\sin (5x-\alpha )+c$,then $\alpha =$

• A. ${\tan }^{-1}\frac{5}{7}$
• B. ${\tan }^{-1}\frac{7}{5}$
• C. ${\tan }^{-1}\frac{1}{5}$
• D. ${\tan }^{-1}\frac{1}{7}$

Answer 1

Answer: (A)

${\tan }^{-1}\frac{5}{7}$

We'll apply integration by parts,

$I=\int e^{ax}\sin bxdx$

$I=e^{ax}\int \sin bxdx-\int a{e}^{ax}\left(\int \sin bxdx\right)dx$

$I=e^{ax}\left[\frac{-\cos bx}{b}\right]-\int a{e}^{ax}\left(\frac{-\cos bx}{b}\right)dx$

$I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}\int e^{ax}\cos bxdx$

Applying integration by parts again,

$I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}\left[e^{ax}\int \cos bxdx-\int a{e}^{ax}\left(\int \cos bxdx\right)dx\right]$

$I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}[e^{ax}\left[\frac{\sin bx}{b}\right]-\int a{e}^{ax}\left(\frac{\sin bx}{b}\right)dx]$

$I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}[\frac{e^{ax}\sin bx}{b}-\frac{a}{b}\int e^{ax}\sin bxdx]$

But we know that,

$I=\int e^{ax}\sin bxdx$

So,

$I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}[\frac{e^{ax}\sin bx}{b}-\frac{a}{b}I]I=\frac{-e^{ax}\cos bx}{b}+\frac{a{e}^{ax}\sin bx}{b^2}-\frac{a^2}{b^2}I(1+\frac{a^2}{b^2})I=\frac{e^{ax}}{b^2}[a\sin bx-b\cos bx]\therefore I=\frac{e^{ax}}{a^2+b^2}[a\sin bx-b\cos bx]+c$

Let us remember the above formula.

Now simplifying the above given question,

$I=\int e^{7x}\sin 5xdx=\frac{e^{7x}}{\sqrt{74}}\sin (5x-\alpha )+c=\frac{e^{7x}}{\sqrt{74}}[\cos \alpha \sin 5x-\sin \alpha \cos 5x]+c$....(1)

Also, by using the calculated formula,

$I=\frac{e^{7x}}{74}[7\sin 5x-5\cos 5x]+c$

$I=\frac{e^{7x}}{\sqrt{74}}[\frac{7}{\sqrt{74}}\sin 5x-\frac{5}{\sqrt{74}}\cos 5x]+c$....(2)

Now comparing, 1 & 2

$\frac{e^{7x}}{\sqrt{74}}[\frac{7}{\sqrt{74}}\sin 5x-\frac{5}{\sqrt{74}}\cos 5x]+c=\frac{e^{7x}}{\sqrt{74}}[\cos \alpha \sin 5x-\sin \alpha \cos 5x]+c$

By comparing we get,

$\sin \alpha =\frac{5}{\sqrt{74}},\cos \alpha =\frac{7}{\sqrt{74}}\therefore \tan \alpha =\frac{5}{7},\alpha ={\tan }^{-1}\frac{5}{7}$