Get the solution to "Evaluate the integral. ∫3sec2t6+3tan⁡tdt"

Ramsey

Answered question

2021-11-07

Evaluate the integral.

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t

mhalmantus

Skilled2021-11-08Added 105 answers

Step 1
Given integral is,

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t

Step 2
To evaluate the given integral.
Given integral is

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t = \int \frac{3}{6} \sec^{2} t d t + \int 3 \tan t d t

= \frac{1}{2} \int \sec^{2} t d t + 3 \int \tan t d t

Now we know

\int \sec^{2} t d t = \tan t + C

To find

\int \tan t d t

Rewrite

\tan t a s \tan t = \frac{\sin t}{\cos t}

Substitute

\cos t = u \Rightarrow - \sin t d t = d u

Then the integral becomes,

\int \tan t d t = \int \frac{\sin t}{\cos t} d t

= \int \frac{- 1}{u} d u

= - \ln | u |

Resubstitute

u = \cos t

again we get,

\Rightarrow \int \tan t d t = - \ln | \cos t | + C

Then the integral becomes,

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t = \frac{1}{2} \tan t - 3 \ln | \cos t | + C

Step 3
Therefore

\int \frac{3 \sec^{2} t}{6} + 3 \tan t d t = \frac{1}{2} \tan t - 3 \ln | \cos t | + C

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