Question

The value of the integral $\int \frac{co{s}^{3}x+co{s}^{5}x}{si{n}^{2}x+si{n}^{4}x}$ dx is

• A. $sinx-6ta{n}^{-1}\left(sinx\right)+c$
• B. $sinx-2(sinx{)}^{-1}+c$
• C. $sinx-2(sinx{)}^{-1}-6ta{n}^{-1}(sinx)+c$
• D. $sinx-2(sinx{)}^{-1}+5ta{n}^{-1}(sinx)+c$

Answer 1

The correct option is C $sinx-2(sinx{)}^{-1}-6ta{n}^{-1}(sinx)+c$

Let I = $\int \frac{co{s}^{3}x+co{s}^{5}x}{si{n}^{2}x+si{n}^{4}x}dx=\int \frac{(co{s}^{2}x+co{s}^{4}x)cosx}{si{n}^{2}x(1+si{n}^{2}x)}dx=\int \frac{[1-si{n}^{2}x+(1-si{n}^{2}x{)}^2]cosx}{si{n}^{2}x(1+si{n}^{2}x)}dx=\int \frac{(2-3si{n}^{2}x+si{n}^4)cosx}{si{n}^{2}x(1+si{n}^{2}x)}dx\text{put}sinx=t\Rightarrow cosxdx=dtI=\int \frac{2-3t^2+t^4}{t^4+t^2}dt\Rightarrow I=\int (1+\frac{2}{t^2}-\frac{6}{t^2+1})dt=t-\frac{2}{t}-6ta{n}^{-1}\left(t\right)+c=sinx-2(sinx{)}^{-1}-6ta{n}^{-1}(sinx)+c$