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Antiderivatives: Inverse Trigonometric Functions

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Wize University Calculus 1 Textbook > Integrals

Antiderivatives of Inverse Trig Functions

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Evaluate dducos⁡−1usin⁡−1u\displaystyle \frac{\text{d}}{\text{d}u} \frac{\cos^{-1}u}{\sin^{-1}u}. It is not necessary to simplify.
More Antiderivatives of Inverse Trig Functions Questions:
Integration Medley
Practice Question
Evaluate ∫x4+1x2+1  ⁣dx{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}.
Evaluate the integral ∫x4+1x2+1dx\displaystyle\int\frac{x^4+1}{x^2+1}dx.
Find the value of the definite integral ∫π/6π/3(sec⁡x+tan⁡x)sec⁡xdx\displaystyle \int_{\pi/6 }^{\pi/3}(\sec x+\tan x)\sec x dx .
Evaluate the indefinite integral ∫6x+e−6x−3x2+1dx\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx.
Evaluate the indefinite integral ∫6x+e−6x−3x2+1dx\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx.
Practice: Definite Integral (~F2017 Final Q26)
Practice: Definite Integral
Evaluate ∫0319+x2+2xln⁡2 dx\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx.
Integration Medley
Practice Question
Evaluate ∫x4+1x2+1  ⁣dx{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}.
Integration Medley
Practice Question
Evaluate ∫x4+1x2+1  ⁣dx{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}.
Integration Medley
Practice Question
Evaluate ∫x4+1x2+1  ⁣dx{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}.
Evaluate ∫−10dx1−x2\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
Integration Medley
Practice Question
Evaluate ∫x4+1x2+1  ⁣dx{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}.
Practice: Definite Integral
Evaluate ∫0319+x2+2xln⁡2 dx\displaystyle \int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx.
Evaluate the integral ∫x4+1x2+1dx\displaystyle\int\frac{x^4+1}{x^2+1}dx.
Find the value of the definite integral ∫π/6π/3(sec⁡x+tan⁡x)sec⁡xdx\displaystyle \int_{\pi/6 }^{\pi/3}(\sec x+\tan x)\sec x dx .
Find the indefinite integral ∫(11+x+11+x2)dx\int(\frac{1}{1+x}+\frac{1}{1+x^2})dx.
Evaluate the indefinite integral ∫6x+e−6x−3x2+1dx\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx.
Evaluate ∫−10dx1−x2\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
Antiderivatives: Inverse Trigonometric Functions
Find ∫24−t2dt\int_{ }^{ }\frac{2}{\sqrt{4-t^2}}dt.
Practice: Definite Integral
Practice: Definite Integral
Evaluate ∫0319+x2+2xln⁡2 dx\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx.