A) \[{{\log }_{e}}\,\left| \sec \,\left( \frac{{{x}^{2}}-1}{2} \right) \right|\,\,+c\]
B) \[{{\log }_{e}}\,\left| \frac{1}{2}{{\sec }^{2}}\,({{x}^{2}}-1) \right|\,\,+c\]
C) \[\frac{1}{2}\,{{\log }_{e}}\,\left| {{\sec }^{2}}\,\left( \frac{{{x}^{2}}-1}{2} \right) \right|\,\,+c\]
D) \[\frac{1}{2}\,{{\log }_{e}}\,\left| \sec ({{x}^{2}}-1) \right|\,\,+\,\,c\]
Correct Answer: A
Solution :
\[\int{x\,\sqrt{\frac{2\,\sin \,({{x}^{2}}-1)\,\,-\,sin\,2\,({{x}^{2}}\,-\,1)}{2\,\sin \,({{x}^{2}}-1)\,+\,sin\,2\,({{x}^{2}}-1)}}}dx\] \[=\,\,\,\int{x\,\sqrt{\frac{2\,\sin \,({{x}^{2}}-1)\,\,-\,sin\,2\,({{x}^{2}}\,-\,1)cos\,({{x}^{2}}-1)}{2\,\sin \,({{x}^{2}}-1)\,+\,sin\,2\,({{x}^{2}}-1)\,cos\,({{x}^{2}}-1)}}}dx\] \[=\,\,\,\int{x\,\sqrt{\frac{1-\cos \,({{x}^{2}}-1)}{1+\cos \,({{x}^{2}}-1)}}}dx\] \[=\,\,\,\int{x\,\sqrt{\frac{2\,{{\sin }^{2}}\,\left( \frac{{{x}^{2}}\,-1}{2} \right)}{2{{\cos }^{2}}\,(\frac{{{x}^{2}}\,-1}{2})}}}dx\] \[=\,\,\,\int{x\,\,\tan \,\left( \frac{{{x}^{2}}-1}{2} \right)}\,dx\] Put \[\frac{{{x}^{2}}-1}{2}\,\,=\,\,t\] \[\Rightarrow \,\,\,\,{{x}^{2}}\,-1\,\,=2t\] \[\Rightarrow \,\,\,2x\,\,dx\,\,=\,\,2dt\] \[\therefore \,\,\int{ta}n\,(t)\,dt\,\,=\,\,ln\,\,\left| \sec \,t \right|\,\,+\,c\] \[=\,\,\,\,\ln \,\left| \sec \,\left( \frac{{{x}^{2}}-1}{2} \right) \right|\,\,+c\]
You need to login to perform this action.
You will be redirected in
3 sec
