A) \[-\cot x+\frac{{{\cot }^{3}}x}{3}+c\]
B) \[\frac{{{\cot }^{3}}x}{3}-\cot x+c\]
C) \[-\cot x-\frac{{{\cot }^{3}}x}{3}+c\]
D) \[\cot x-\frac{{{\cot }^{2}}x}{3}+c\]
Correct Answer: C
Solution :
Let \[I=\int{\cos {{\sec }^{4}}x}dx\] \[=\int{\cos {{\sec }^{2}}x(1+{{\cot }^{2}}x)}dx\] \[=\int{\cos {{\sec }^{2}}xdx+\int{\cos e{{c}^{2}}x{{\cot }^{2}}x}}dx\] Let \[cot\text{ }x=t\] \[\Rightarrow \] \[-cose{{c}^{2}}x\text{ }dx=dt\] \[\Rightarrow \] \[cose{{c}^{2}}x\,dx=-dt\] \[\therefore \] \[I=-\cot x-\int{{{t}^{2}}dt}\] \[=-\cot x-\frac{{{t}^{3}}}{3}+c\] \[=-\cot x-\frac{{{\cot }^{3}}x}{3}+c\]
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