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Question Number 215687 by Ajeemkhan last updated on 15/Jan/25

Answered by som(math1967) last updated on 15/Jan/25

$$\int {sec}^{p-1}xsecxtanxdx$$$$\Rightarrow \int {sec}^{p-1}xd\left(secx\right)$$$$\Rightarrow \frac{1}{p}{sec}^{p}x+C$$

Commented by Ajeemkhan last updated on 15/Jan/25

$$\mathrm{Answer}\mathrm{is}\mathrm{right}\mathrm{but}\mathrm{method}\mathrm{not}\mathrm{understood}$$

Commented by Ajeemkhan last updated on 15/Jan/25

$$\mathrm{Answer}\mathrm{is}\mathrm{right}\mathrm{but}\mathrm{method}\mathrm{not}\mathrm{understood}$$

Commented by som(math1967) last updated on 15/Jan/25

$$\int \sec {}^{p-1}xsecxtanxdx$$$$letsecx=t\Rightarrow secxtanxdx=dt$$$$\therefore \int t^{p-1}dt=\frac{1}{p}{t}^p+C=\frac{1}{p}\sec {}^{p}x+C$$

Answered by MATHEMATICSAM last updated on 15/Jan/25

$$\mathrm{Put}\sec x=t.$$$$\Rightarrow \sec x\tan xdx=dt$$$$$$$$\int {\sec }^{p}x\tan xdx$$$$=\int {\sec }^{p-1}x.\sec x\tan xdx$$$$=\int t^{p-1}dt$$$$=\frac{t^p}{p}+\mathrm{C}$$$$=\frac{{\sec }^{p}x}{p}+\mathrm{C}$$

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