5.4 Indefinite Integrals and the Net Change Theorem/17: Difference between revisions

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17)<math>\int_{}^{}1+tan^2xdx</math> =
<math>
\int(1+\tan^2{\alpha})\,d\alpha = \int\sec^2\alpha \,d\alpha = \tan\alpha + C
</math>


<math>\int_{}^{}1+\frac{sin^2x}{cos^2x}dx</math> =


<math>\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx</math>  
Note: <math>1+\tan^2{\alpha} = \sec^2\alpha</math>


<math>\cos^2x+sin^2x=1</math> thus,


<math>\int_{}^{}\frac{1}{cos^2x}dx</math> =
Or,


<math>\int_{}^{}\sec^2xdx</math> =


<math>tanx+C</math>
<math>
 
\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha  = \int\frac{1}{cos^2\alpha}\,d\alpha =
 
\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C
</math>
 
 
Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

Latest revision as of 19:39, 21 September 2022


Note:


Or,



Note:

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