Question

$\int e^{-3x}\cos 7xdx=\frac{e^{3x}}{\sqrt{58}}[\sin \alpha \cos 7x+\cos \alpha \sin 7x]+c$, then values of $\sin \alpha$ and $\cos \alpha$ are

• A. $\frac{3}{\sqrt{58}},\frac{7}{\sqrt{58}}$
• B. $\frac{-3}{\sqrt{58}},\frac{7}{\sqrt{58}}$
• C. $\frac{-3}{58},\frac{7}{58}$
• D. $\frac{-7}{\sqrt{58}},\frac{3}{\sqrt{58}}$

Answer 1

Answer: (B)

$\frac{-3}{\sqrt{58}},\frac{7}{\sqrt{58}}$

Applying integration by parts,

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

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

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

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

Let's use integration by parts again,

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

But we know that,

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

So,

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

Now $a=-3$ , $b=7$

$\therefore I=\frac{e^{-3x}}{(-3{)}^2+7^2}[-3\cos 7x+7\sin 7x]+cI=\frac{e^{-3x}}{58}[-3\cos 7x+7\sin 7x]+cI=\frac{e^{-3x}}{\sqrt{58}}[\frac{-3}{\sqrt{58}}\cos 7x+\frac{7}{\sqrt{58}}\sin 7x]+c$

Hence by comparing,

$\sin \alpha =\frac{-3}{\sqrt{58}},\cos \alpha =\frac{7}{\sqrt{58}}$.