Let $I_{n}=\int an ^{n} x d x,(n1)$. If $I_{4}+I_{6}=a an ^{5} x+b x^{5}+C$, where $C$ is a con

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Let $I_{n}=\int 	an ^{n} x d x,(n>1)$. If $I_{4}+I_{6}=a 	an ^{5} x+b x^{5}+C$, where $C$ is a constant of integration, then the ordered pair $(a, b)$ is equal to\

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(a) : We have $I_{n}=\int 	an ^{n} x d x,(n>1)$\ $=\int 	an ^{n-2} x\left(\sec ^{2} x-1ight) d x$\ $=\int 	an ^{n-2} x \sec ^{2} x d x-\int 	an ^{n-2} x d x$\ $=\frac{	an ^{n-1} x}{n-1}-I_{n-2}+C$\ Then, $I_{n}+I_{n-2}=\frac{(	an x)^{n-1}}{n-1}+C$\ Now, $I_{4}+I_{6}=\frac{	an x^{5}}{5}+C$\ And $I_{4}+I_{6}=a 	an x^{5}+b x^{5}+C$ (Given) ...(ii) On comparing (i) and (ii), we get\ $a=\frac{1}{5}, b=0$,\ and $C$ is a constant of integration.\