Question

Evaluate: $\int \frac{1}{1+3si{n}^{2}x+8co{s}^{2}x}dx$

• A. $\frac{1}{6}ta{n}^{-1}\left(2tanx\right)+C$
• B. $ta{n}^{-1}\left(2tanx\right)+C$
• C. $\frac{1}{6}ta{n}^{-1}\left(\frac{2tanx}{3}\right)+C$
• D. None of these

Answer 1

Answer: (C)

$I=\int \frac{1}{1+3si{n}^{2}x+8co{s}^{2}x}dx$

Dividing the numerator and denominator by$co{s}^{2}x$, we get

$\Rightarrow I=\int \frac{se{c}^{2}x}{se{c}^{2}x+3ta{n}^{2}x+8}dx$

$\Rightarrow I=\int \frac{se{c}^{2}x}{1+ta{n}^{2}x+3ta{n}^{2}x+8}dx=\int \frac{se{c}^{2}x}{4ta{n}^{2}x+9}dx$

putting $tanx=t\Rightarrow se{c}^{2}xdx=dt$, we get

$I=\int \frac{dt}{4t^2+9}=\frac{1}{4}\int \frac{dt}{t^2+{\left(\frac{3}{2}\right)}^2}=\frac{1}{4}\times \frac{1}{\frac{3}{2}}ta{n}^{-1}\left(\frac{t}{\frac{3}{2}}\right)+C$

$\Rightarrow I=\frac{1}{6}ta{n}^{-1}\left(\frac{2t}{3}\right)+C=\frac{1}{6}ta{n}^{-1}\left(\frac{2tanx}{3}\right)+C$