Question

If ƒ(x) = $\frac{x^{2}}{1 + x^{2}}$ and g(x) = sin x, then $\int_{}^{}{(ƒog)}$ (x) cos (x) dx=

• A. sin x – tan^–1 (sin x) + c
• B. cos x – tan^–1 (sin x) + c
• C. cos x + tan^–1 (sin x) + c
• D. sin x – tan^–1 (cos x) + c

Answer 1

Answer: (A)

sin x – tan^–1 (sin x) + c

Answer 2

I =$\int_{}^{}{(ƒog)}$ (x) cos (x) dx = $\int_{}^{}{(ƒ(g(x)))}$cos x dx

= $\int_{}^{}{ƒ(\sin x)}$ cos x dx = $\int_{}^{}\frac{\sin^{2}x}{1 + \sin^{2}x}$cos x dx.

= $\int_{}^{}{\frac{t^{2}}{1 + t^{2}}dt}$ (where sin x = t Ž cos x dx = dt)

= $\int_{}^{}1$– $\frac{1}{1 + t^{2}}$ dt = t – tan^–1 t + c

= sin t – tan^–1 (sin t) + c.