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Question Number 12533 by tawa last updated on 24/Apr/17

$$\mathrm{compute}$$$$\int {\sec }^5\left(\mathrm{x}\right){\tan }^3\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by sma3l2996 last updated on 25/Apr/17

$$I=\int {sec}^5\left(x\right){tan}^3\left(x\right)dx=\int \frac{{sin}^{3}x}{{cos}^{8}x}dx$$$$u={sin}^{2}x\Rightarrow u^{\prime }=2cosxsinx$$$$v^{\prime }=\frac{sinx}{{cos}^{8}x}\Rightarrow v=\frac{1}{7{cos}^{7}x}$$$$I=\frac{{sin}^{2}x}{7{cos}^{7}x}-\frac{2}{7}\int \frac{sinx}{{cos}^{6}x}dx+c$$$$I=\frac{{sin}^{2}x}{7{cos}^{7}x}-\frac{2}{7}\times \left(\frac{1}{5{cos}^{5}x}\right)+C$$$$=\frac{1}{7}{sec}^5\left(x\right){tan}^2\left(x\right)-\frac{2}{35}{sec}^5\left(x\right)+C$$$$I=\frac{1}{7}{sec}^5\left(x\right)({tan}^2\left(x\right)-\frac{2}{5})+C$$

Commented by tawa last updated on 25/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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