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Integrals Resulting in Inverse Trigonometric Functions.

The table below contains the derivatives of the inverse trigonometric functions that arise frequently in integration.

Izl < a

4 cos

Irl < a

# (Gcot ‘(E)) a? + r2

sec

Iz >

CSC

With help from the table above, evaluate the following indefinite integral: Hint: simple substitution may help.

âˆš64

Note: the answer should be in terms of Î¸ only:

1/tan

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02:12

Integrals Resulting in Inverse Trigonometric FunctionsThe table below contains the derivatives of the inverse trigonometric functions that arise frequently in integration:sin^(-1) x < acos^(-1) x < atan^(-1) x/asqrt(a^2 + x^2)cot^(-1) x/asqrt(a^2 + x^2)sec^(-1) x > acsc^(-1) x > a1/sqrt(x^2 – a^2)With help from the table above, evaluate the following indefinite integral. Hint: a simple substitution may help.âˆ«(1/sqrt(1 + 9e^(2x))) dxNote: answer should be in terms of x only.

05:28

Integrals Resulting in Inverse Trigonometric Functions

The table below contains the derivatives of the inverse trigonometric functions that arise frequently in integration:

sin^(-1)(x) < acos^(-1)(x) < atan^(-1)(x) a^2 + x^2cot^(-1)(x) a^2 + x^2sec^(-1)(x) > acsc^(-1)(x) > a

With help from the table above, evaluate the following indefinite integral. Hint: simple substitution may help.

âˆ«(9 sec^(-1)(âˆš(t^4 – 81))) dt

Note: answer should be in terms of only âˆš(t) and âˆš(t^2)

02:21

The table below contains the derivatives of the inverse trigonometric functions that arise frequently in integration.

sin^(-1)'(x) = 1/âˆš(1 – x^2)

cos^(-1)'(x) = -1/âˆš(1 – x^2)

tan^(-1)'(x) = 1/(1 + x^2)

cot^(-1)'(x) = -1/(1 + x^2)

sec^(-1)'(x) = 1/(|x|âˆš(x^2 – 1))

csc^(-1)'(x) = -1/(|x|âˆš(x^2 – 1))

âˆ«4sec(x) dx = 4ln|sec(x) + tan(x)| + C

Note: the answer should be in terms of x only.

08:11

Evaluate the integrals by making a substitution (possibly trigonometric) and then applying a reduction formula.$$\int \frac{\csc ^{3} \sqrt{\theta}}{\sqrt{\theta}} d \theta$$

Transcript

Hi everyone. Hi everyone. Welcome to this video. So here let us do the integral value with x divided by root of 64 minus x power 4 dx. So here let us write this as x divided by root of 8 squared minus x squared the whole squared. So here we have x divided by x into 8 squared by x squared minus x squared. So let us put root over here. So here x and x will be cancelled. So it will be equal to 8 by x the whole squared minus x squared dx. Now this integral value is…

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