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Math  /  Calculus

Question∫sec⁡(1−z)tan⁡(1−z)dz\int \sec (1-z) \tan (1-z) d z
Please solve for this probl

Studdy Solution

STEP 1

1. The integral involves the functions sec⁡(1−z)\sec(1-z) and tan⁡(1−z)\tan(1-z), which are trigonometric functions.
2. We need to recognize that the integral of sec⁡(u)tan⁡(u)\sec(u) \tan(u) with respect to uu is a standard integral.
3. We'll use a substitution method to simplify the integral.

STEP 2

1. Perform a substitution to simplify the integral.
2. Integrate with respect to the new variable.
3. Substitute back to the original variable.

STEP 3

Let u=1−zu = 1 - z. Then, du=−dzdu = -dz.
dz=−du dz = -du

STEP 4

Substitute u=1−zu = 1 - z and dz=−dudz = -du into the integral.
∫sec⁡(1−z)tan⁡(1−z) dz=∫sec⁡(u)tan⁡(u) (−du) \int \sec(1-z) \tan(1-z) \, dz = \int \sec(u) \tan(u) \, (-du)

STEP 5

Rewrite the integral with the negative sign factored out.
∫sec⁡(u)tan⁡(u) (−du)=−∫sec⁡(u)tan⁡(u) du \int \sec(u) \tan(u) \, (-du) = -\int \sec(u) \tan(u) \, du

STEP 6

Recognize that the integral of sec⁡(u)tan⁡(u)\sec(u) \tan(u) with respect to uu is a standard integral.
∫sec⁡(u)tan⁡(u) du=sec⁡(u)+C \int \sec(u) \tan(u) \, du = \sec(u) + C

STEP 7

Apply the result to our integral.
−∫sec⁡(u)tan⁡(u) du=−sec⁡(u)+C -\int \sec(u) \tan(u) \, du = -\sec(u) + C

STEP 8

Substitute back u=1−zu = 1 - z.
−sec⁡(u)+C=−sec⁡(1−z)+C -\sec(u) + C = -\sec(1 - z) + C
Therefore, the solution to the integral is:
∫sec⁡(1−z)tan⁡(1−z) dz=−sec⁡(1−z)+C \int \sec(1-z) \tan(1-z) \, dz = -\sec(1 - z) + C

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