Question

$\int_{}^{}{\frac{\sin 2x}{a^{2}\sin^{2}x + b^{2}\cos^{2}x}dx}$ is equal to

• A. $\frac{1}{b^{2} - a^{2}}\log(a^{2}\sin^{2}x + b^{2}\cos^{2}x) + c$
• B. $\frac{1}{a^{2} - b^{2}}\log(a^{2}\sin^{2}x + b^{2}\cos^{2}x) + c$
• C. $\log(a^{2}\sin^{2}x - b^{2}\cos^{2}x) + c$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{a^{2} - b^{2}}\log(a^{2}\sin^{2}x + b^{2}\cos^{2}x) + c$

Answer 2

Put $a^{2}\sin^{2}x + b^{2}\cos^{2}x = t$

⇒ $(a^{2}.2\sin x\cos x - b^{2}.2\cos x\sin x)dx = dt$

⇒ $\sin 2x(a^{2} - b^{2})dx = dt$

$\int \frac { \sin 2 x } { a ^ { 2 } \sin ^ { 2 } x + b ^ { 2 } \cos ^ { 2 } x } d x = \frac { 1 } { \left( a ^ { 2 } - b ^ { 2 } \right) }$ $\int_{}^{}{\frac{dt}{t} = \frac{1}{a^{2} - b^{2}}\log t + c} = \frac{1}{a^{2} - b^{2}}\log(a^{2}\sin^{2}x + b^{2}\cos^{2}x) + c$