Evaluate the integrals ∫etcos⁡(3et−2)dt

Question

Nicole Keller · Accepted Answer

Step 1
We have to evaluate the integral:

\int e^{t} \cos (3 e^{t} - 2) d t

In this case we should use substitution method since derivatives of one function is in the integral.
So let

x = 3 e^{t} - 2

differentiating both sides with respect to t, we get

\frac{d x}{d t} = \frac{d (3 e^{t} - 2)}{d t}

= 3 \frac{d e^{t}}{d t} - \frac{d 2}{d t}

= 3 e^{t} - 0

\frac{d x}{d t} = 3 e^{t}

\frac{d x}{3} = e^{t} d t

Step 2
Substituting above values, we get

\int e^{t} \cos (3 e^{t} - 2) d t = \int \cos (3 e^{t} - 2) e^{t} d t

= \frac{1}{3} \int \cos x d x

= \frac{1}{3} (\sin x) + C

Where, C is an arbitrary constant.
Hence, value of given integral is

\frac{1}{3} \sin x + C

.

Prioned · Accepted Answer

Step 1: Use Integration by Substitution.
Let

u = 3 e^{t} - 2, d u = 3 e^{t} d t, t h e n e^{t} d t = \frac{1}{3} d u

Step 2: Using u and du above, rewrite

\int e^{t} \cos (3 e^{t} - 2) d t

.

\int \frac{\cos u}{3} d u

Step 3: Use Constant Factor Rule:

\int c f (x) d x = c \int f (x) d x

.

\frac{1}{3} \int \cos u d u

Step 4: Use Trigonometric Integration: the integral of

\cos u i s \sin u

.

\frac{\sin u}{3}

Step 5: Substitute

u = 3 e^{t} - 2

back into the original integral.

\frac{\sin (3 e^{t} - 2)}{3}

Step 6: Add constant.

\frac{\sin (3 e^{t} - 2)}{3} + C

Evaluate the integrals \int e^{t}\cos (3e^{t}-2)dt