A) \[8\,\left( \sin \frac{x}{8}+\cos \frac{x}{8} \right)+C\]
B) \[8\,\left( \sin \frac{x}{8}-\cos \frac{x}{8} \right)+C\]
C) \[8\,\left( \cos \frac{x}{8}-\sin \frac{x}{8} \right)+C\]
D) \[8\,\left( \sin \frac{x}{8}-\cos \frac{x}{8} \right)+C\]
Correct Answer: B
Solution :
We have \[\int{\sqrt{1+\sin \left( \frac{x}{4} \right)d\,x}}\] \[=\int{\sqrt{1+2\sin \frac{x}{8}{{\cos }^{2}}\frac{x}{8}}\,dx}\] \[[\sin 2x=2\sin x\cos x]\] \[=\int{\sqrt{{{\sin }^{2}}+\frac{x}{8}+{{\cos }^{2}}\frac{x}{8}+2\sin \frac{x}{8}\cos \frac{x}{8}}\,dx}\] \[[1={{\sin }^{2}}x+{{\cos }^{2}}x]\] \[=\int{\sqrt{{{\left( \sin +\frac{x}{8}+\cos \frac{x}{8} \right)}^{2}}}\,dx}\] \[[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] \[=\int{\left( \sin \frac{x}{8}\cos \frac{x}{8} \right)dx}\] \[=\frac{-\cos \frac{x}{8}}{1/8}+\frac{\sin \frac{x}{8}}{1/8}+C\] \[=8\left( -\cos \frac{x}{8}+\sin \frac{x}{8} \right)+C\] \[=8\left( \sin \frac{x}{8}-\cos \frac{x}{8} \right)+C\]
You need to login to perform this action.
You will be redirected in
3 sec
