question_answer

Evaluate \[\int{\sqrt[3]{\frac{{{\sin }^{2}}x}{{{\cos }^{14}}x}}dx.}\]

Answer:

Let \[l=\int{\sqrt[3]{\frac{{{\sin }^{2}}x}{{{\cos }^{14}}x}}dx.}\]                         \[=\int{{{\sin }^{2/3}}x.co{{s}^{-14/3}}x\,dx}\] Here, the sum of exponents of \[\sin x\] and \[co\operatorname{s}x\]is \[-\,4,\] which is a negative even integer. So, on dividing and multiplying by \[{{\cos }^{4}}x,\] we get \[l=\int{{{\sin }^{2/3}}x\cdot co{{s}^{-14/3}}x\cdot \frac{{{\cos }^{4}}x}{{{\cos }^{4}}x}\,dx}\] \[=\int{\frac{{{\sin }^{2/3}}x}{co{{s}^{2/3x}}x}\cdot se{{c}^{4}}x\,dx}\] \[=\int{{{\tan }^{2/3}}x(1+{{\tan }^{2}}x){{\sec }^{2}}x\,dx}\] \[[\because \,\,\,{{\sec }^{2}}x=1+{{\tan }^{2}}x]\] Put \[\tan \,x=t\] \[\Rightarrow \] \[{{\sec }^{2}}x\,dx=dt\] \[\therefore \]      \[l=\int{{{t}^{2/3}}(1+{{t}^{2}})\,dt}\]             \[=\int{({{t}^{2/3}}+{{t}^{8/3}})dt}=\frac{3}{5}{{t}^{5/3}}+\frac{3}{11}{{t}^{11/3}}+C\]             \[=\frac{3}{5}{{\tan }^{5/3}}x+\frac{3}{11}{{\tan }^{11/3}}x+C\] [put\[t=\tan x\]]