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Question Number 59563 by aliesam last updated on 11/May/19

Commented by aliesam last updated on 11/May/19

$$thankyousir$$

Answered by MJS last updated on 12/May/19

$$\int \frac{{\sin }^{-7}x}{{\cos }^{3}x}dx=\int \frac{dx}{{\sin }^{7}x{\cos }^{3}x}=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{dt}{t^7{(1-t^2)}^2}=\int (\frac{1}{4{(1-t)}^2}-\frac{1}{4{(1+t)}^2}+\frac{2}{1-t}-\frac{2}{1+t}+\frac{4}{t}+\frac{3}{t^3}+\frac{2}{t^5}+\frac{1}{t^7})dt$$$$\mathrm{and}\mathrm{now}\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{easy}$$

Commented by malwaan last updated on 11/May/19

$$butyoursolutionisveryshort$$$$Canyoumakestepsplease?$$

Answered by tanmay last updated on 12/May/19

$$\int \frac{dx}{{sin}^7{xcos}^{3}x}$$$$\int \frac{dx}{{tan}^{7}x\times {cos}^{10}x}$$$$\int \frac{{sec}^{2}x\times {sec}^{8}x}{{tan}^{7}x}dx$$$$\int \frac{{(1+t^2)}^{4}dt}{t^7}[t=tanx]$$$$\int \frac{1+4c_1{t}^2+4c_2{t}^4+4c_3{t}^6+t^8}{t^7}dt$$$$\int t^{-7}dt+4\int t^{-5}dt+6\int t^{-3}+4\int \frac{dt}{t}+\int tdt$$$$\frac{t^{-6}}{-6}+4\times \frac{t^{-4}}{-4}+6\times \frac{t^{-2}}{-2}+4lnt+\frac{t^2}{2}+c$$$$\frac{-1}{6{\left(tanx\right)}^6}-\frac{1}{{\left(tanx\right)}^4}-\frac{3}{{\left(tanx\right)}^2}+4ln\left(tanx\right)+\frac{{\left(tanx\right)}^2}{2}+c$$

Commented by malwaan last updated on 12/May/19

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}$$

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