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Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
2 sec^2(t)/tan^2(t) + 14 tan(t) + 48 dt

Sagot :

LammettHashidn

Complete the square in the denominator to get

tan²(t ) + 14 tan(t ) + 48 = tan²(t ) + 14 tan(t ) + 49 - 1

… = (tan(t ) + 7)² - 1

Substitute u = tan(t ) + 7 and du = sec²(t ) dt. Then the integral becomes

[tex]\displaystyle \int \frac{2\sec^2(t)}{\tan^2(t)+14\tan(t)+48} \,\mathrm dt = 2 \int \frac{\mathrm du}{u^2-1}[/tex]

Separate the integrand into partial fractions:

[tex]\dfrac1{u^2-1} = \dfrac12 \left(\dfrac1{u-1}-\dfrac1{u+1}\right)[/tex]

Then we get

[tex]\displaystyle \int \frac{2\sec^2(t)}{\tan^2(t)+14\tan(t)+48} \,\mathrm dt =  \int \left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm du \\\\ =\ln|u-1|-\ln|u+1| + C \\\\ = \ln\left|\frac{u-1}{u+1}\right|+C \\\\ = \boxed{\ln\left|\frac{\tan(t)+6}{\tan(t)+8}\right|+C}[/tex]

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