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Question Number 96613 by s.ayeni14@yahoo.com last updated on 03/Jun/20

$$\int \sec \theta \mathrm{d}\theta$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int sec\left(\theta \right)\frac{sec\left(\theta \right)+tan\left(\theta \right)}{sec\left(\theta \right)+tan\left(\theta \right)}d\theta$$$$=ln\mid sec\left(\theta \right)+tan\left(\theta \right)\mid +c$$$$ihavemorethan9waystosolvethis$$$$integraliwillpostthemlater$$

Commented by s.ayeni14@yahoo.com last updated on 03/Jun/20

$$\mathrm{ok}\mathrm{sir}$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$=\int sec\left(x\right)+tan\left(x\right)-tan\left(x\right)dx$$$$=\int \frac{1-sin\left(x\right)}{cos\left(x\right)}dx+\int tan\left(x\right)dx$$$$=\int \frac{cos\left(x\right)}{1+sin\left(x\right)}dx+\int \frac{sin\left(x\right)}{cos\left(x\right)}dx$$$$ln\mid 1+sin\left(x\right)\mid -ln\mid cos\left(x\right)\mid +c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$=\int \frac{1}{sin(\frac{\pi }{2}-x)}dx$$$$=\int \frac{1}{2sin(\frac{\pi }{4}-x)cos(\frac{\pi }{4}-x)}.\frac{{sec}^2(\frac{\pi }{4}-\frac{x}{2})}{{sec}^2(\frac{\pi }{4}-\frac{x}{2})}dx$$$$=\frac{1}{2}\int \frac{{sec}^2(\frac{\pi }{4}-\frac{x}{2})}{tan(\frac{\pi }{4}-\frac{x}{2})}dx$$$$=-ln\mid tan(\frac{\pi }{4}-\frac{x}{2})\mid +c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int sec\left(x\right)+tan\left(x\right)-tan\left(x\right)dx$$$$\int \frac{1+sin\left(x\right)}{cos\left(x\right)}dx-\int tan\left(x\right)dx$$$$\int \frac{{(cos\frac{x}{2}+sin\frac{x}{2})}^2}{{cos}^2\left(\frac{x}{2}\right)-{sin}^2\left(\frac{x}{2}\right)}dx+ln\mid cos\left(x\right)\mid$$$$=-2ln\mid cos\left(\frac{x}{2}\right)-\frac{x}{2}\mid +ln\mid cos\left(x\right)\mid +c$$$$$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$=\int sec\left(x\right)\frac{tan\left(x\right)}{tan\left(x\right)}dx$$$$=\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{{sec}^2\left(x\right)-1}}dx$$$$={cosh}^{-1}\left(sec\left(x\right)\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{sin\left(x\right)}{sin\left(x\right)cos\left(x\right)}dx$$$$\int \frac{sin\left(x\right)}{cos\left(x\right)\sqrt{1-{cos}^2\left(x\right)}}dx$$$$={sech}^{-1}\left(cos\left(x\right)\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{csc\left(x\right)}{cot\left(x\right)}dx$$$$\int \frac{csc\left(x\right)cot\left(x\right)}{cot\left(x\right)}dx$$$$\int \frac{-1}{{csc}^2\left(x\right)-1}d\left(csc\left(x\right)\right)$$$${coth}^{-1}\left(csc\left(x\right)\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{1}{{cos}^2\left(\frac{x}{2}\right)-{sin}^2\left(\frac{x}{2}\right)}.\frac{{sec}^2\left(\frac{x}{2}\right)}{{sec}^2\left(\frac{x}{2}\right)}dx$$$$\int \frac{{sec}^2\left(\frac{x}{2}\right)}{1-{tan}^2\left(\frac{x}{2}\right)}dx$$$$2\int \frac{-\frac{1}{2}{sec}^2\left(\frac{x}{2}\right)}{{tan}^2\left(\frac{x}{2}\right)-1}dx$$$$=2{coth}^{-1}\left(tan\left(\frac{x}{2}\right)\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{csc\left(x\right)}{cot\left(x\right)}dx$$$$\int \frac{{csc}^2\left(x\right)}{csc\left(x\right)cot\left(x\right)}dx$$$$\int \frac{{csc}^2\left(x\right)}{cot\left(x\right)\sqrt{{cot}^2\left(x\right)+1}}dx$$$${csch}^{-1}\left(cot\left(x\right)\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{{sec}^2\left(\frac{x}{2}\right)}{1-{tan}^2\left(\frac{x}{2}\right)}dx$$$$\int \frac{tan\left(\frac{x}{2}\right){sec}^2\left(\frac{x}{2}\right)}{{\left(\sqrt{1-{tan}^2\left(\frac{x}{2}\right)}\right)}^{2}tan\frac{x}{2}}dx$$$$\int \frac{1}{\sqrt{1-{tan}^2\left(\frac{x}{2}\right)}\sqrt{1-{\left(\sqrt{1-{tan}^2\frac{x}{2}}\right)}^2}}.\frac{2tan\frac{x}{2}{sec}^2\frac{x}{2}}{2\sqrt{1-{tan}^2\frac{x}{2}}}dx$$$$=2{sech}^{-1}\sqrt{1-{tan}^2\left(\frac{x}{2}\right)}+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{{sec}^2\left(x\right)-1}}dx=\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{sec\left(x\right)-1}\sqrt{sec\left(x\right)+1}}dx$$$$=\frac{1}{2}\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{\frac{sec\left(x\right)-1}{2}}\sqrt{\frac{sec\left(x\right)+1}{2}}}dx$$$$=2\int \frac{1}{\sqrt{{\left(\sqrt{\frac{sec\left(x\right)+1}{2}}\right)}^2-1}}.\frac{\frac{sec\left(x\right)tan\left(x\right)}{2}}{2\sqrt{\frac{sec\left(x\right)+1}{2}}}$$$$=2{cosh}^{-1}\left(\sqrt{\frac{sec\left(x\right)+1}{2}}\right)+c$$

Answered by  M±th+et+s last updated on 03/Jun/20

$$\int sec\left(x\right)dx=\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{{sec}^2\left(x\right)-1}}dx$$$$=\frac{1}{2}\int \frac{sec\left(x\right)tan\left(x\right)}{\sqrt{\frac{sec\left(x\right)-1}{2}}\sqrt{\frac{sec\left(x\right)+1}{2}}}dx$$$$=\int \frac{1}{\sqrt{\frac{sec\left(x\right)-1}{2}+1}}.\frac{sec\left(x\right)tan\left(x\right)}{2\sqrt{\frac{sec\left(x\right)-1}{2}}}dx$$$$=2{sinh}^{-}\left(\sqrt{\frac{sec\left(x\right)-1}{2}}\right)+c$$$$$$$$andmybethereareanotherways$$

Answered by Rio Michael last updated on 03/Jun/20

$$\mathrm{hahahaha}\mathrm{great}\mathrm{work}\mathrm{sir}!$$$$\mathrm{but}\mathrm{one}\mathrm{method}\mathrm{left}.$$$$\int \sec \theta d\theta =\int \frac{\sec \theta (\sec \theta +\tan \theta )}{\sec \theta +\tan \theta }d\theta =\int \frac{{\sec }^2\theta +\sec \theta \tan \theta }{\sec \theta +\tan \theta }d\theta$$$$\mathrm{let}u=\sec \theta +\tan \theta \Rightarrow du=(\sec \theta \tan \theta +{\sec }^2\theta )d\theta =({\sec }^2\theta +\sec \theta \tan \theta )d\theta$$$$\Rightarrow \int \sec \theta d\theta =\int \frac{du}{u}\mathrm{after}\mathrm{substitution}=\ln \mid u\mid +k$$$$\Rightarrow \int \sec \theta d\theta =\ln \mid \sec \theta +\tan \theta \mid +k$$$$$$

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