Calculus Integrals with inverse trig functions With the inverse trig functions a r c t a n x arctan x a r c t an x a r c s i n x arcsin x a r cs in x
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Inverse trig functions and their derivatives y = arcsin x y = \arcsin x y = arcsin x x = sin y x = \sin y x = sin y ( − π 2 ≤ y ≤ π 2 ) \left(-\frac{\pi}{2} \le y \le \frac{\pi}{2}\right) ( − 2 π ≤ y ≤ 2 π ) y = arctan x y = \arctan x y = arctan x x = tan y x = \tan y x = tan y ( − π 2 < y < π 2 ) \left(-\frac{\pi}{2} < y < \frac{\pi}{2}\right) ( − 2 π < y < 2 π ) sin − 1 x \sin^{-1}x sin − 1 x tan − 1 x \tan^{-1}x tan − 1 x
( arcsin x ) ′ = 1 1 − x 2 , ( arctan x ) ′ = 1 1 + x 2 (\arcsin x)' = \frac{1}{\sqrt{1-x^2}},\qquad (\arctan x)' = \frac{1}{1+x^2} ( arcsin x ) ′ = 1 − x 2 1 , ( arctan x ) ′ = 1 + x 2 1 Complete the square to use the formulas If the denominator is a quadratic that doesn't factor, like x 2 + 2 x + 5 x^2 + 2x + 5 x 2 + 2 x + 5 ( x + 1 ) 2 + 2 2 (x+1)^2 + 2^2 ( x + 1 ) 2 + 2 2 arctan \arctan arctan
∫ d x x 2 + 2 x + 5 = ∫ d x ( x + 1 ) 2 + 2 2 = 1 2 arctan x + 1 2 + C \int \frac{dx}{x^2 + 2x + 5} = \int \frac{dx}{(x+1)^2 + 2^2} = \frac{1}{2}\arctan\frac{x+1}{2} + C ∫ x 2 + 2 x + 5 d x = ∫ ( x + 1 ) 2 + 2 2 d x = 2 1 arctan 2 x + 1 + C If the numerator contains x x x log \log log arctan \arctan arctan
Integrating the inverse trig functions themselves Integrate arctan x \arctan x arctan x arcsin x \arcsin x arcsin x 1 ⋅ arctan x 1 \cdot \arctan x 1 ⋅ arctan x
∫ arctan x d x = x arctan x − ∫ x 1 + x 2 d x = x arctan x − 1 2 log ( 1 + x 2 ) + C \int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx = x\arctan x - \frac{1}{2}\log(1+x^2) + C ∫ arctan x d x = x arctan x − ∫ 1 + x 2 x d x = x arctan x − 2 1 log ( 1 + x 2 ) + C Worked examples
Find the indefinite integral.
∫ 1 x 2 + 16 d x \int \frac{1}{x^{2} + 16}\,dx ∫ x 2 + 16 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint An antiderivative of 1 x 2 + a 2 \frac{1}{x^2 + a^2} x 2 + a 2 1 1 a arctan x a \frac{1}{a}\arctan\frac{x}{a} a 1 arctan a x x = a tan θ x = a\tan\theta x = a tan θ
Answer ∫ 1 x 2 + 16 d x = 1 4 arctan x 4 + C \int \frac{1}{x^{2} + 16}\,dx = \frac{1}{4}\arctan\frac{x}{4} + C ∫ x 2 + 16 1 d x = 4 1 arctan 4 x + C (C C C
Solution Use the formula ∫ d x x 2 + a 2 = 1 a arctan x a + C \displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C ∫ x 2 + a 2 d x = a 1 arctan a x + C a > 0 a > 0 a > 0
∫ 1 x 2 + 16 d x = 1 4 arctan x 4 + C \begin{aligned}\int \frac{1}{x^{2} + 16}\,dx &= \frac{1}{4}\arctan\frac{x}{4} + C\end{aligned} ∫ x 2 + 16 1 d x = 4 1 arctan 4 x + C
Find the indefinite integral.
∫ 1 x 2 − 6 x + 10 d x \int \frac{1}{x^{2} - 6x + 10}\,dx ∫ x 2 − 6 x + 10 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Complete the square in the denominator to get the form ( x + p ) 2 + a 2 (x + p)^2 + a^2 ( x + p ) 2 + a 2
Answer ∫ 1 x 2 − 6 x + 10 d x = arctan ( x − 3 ) + C \int \frac{1}{x^{2} - 6x + 10}\,dx = \arctan(x - 3) + C ∫ x 2 − 6 x + 10 1 d x = arctan ( x − 3 ) + C (C C C
Solution Completing the square in the denominator, x 2 − 6 x + 10 = ( x − 3 ) 2 + 1 x^{2} - 6x + 10 = (x - 3)^{2} + 1 x 2 − 6 x + 10 = ( x − 3 ) 2 + 1 ∫ d x x 2 + a 2 = 1 a arctan x a + C \displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C ∫ x 2 + a 2 d x = a 1 arctan a x + C a > 0 a > 0 a > 0 x − 3 x - 3 x − 3 x x x
∫ 1 x 2 − 6 x + 10 d x = ∫ 1 ( x − 3 ) 2 + 1 d x = arctan ( x − 3 ) + C \begin{aligned}\int \frac{1}{x^{2} - 6x + 10}\,dx &= \int \frac{1}{(x - 3)^{2} + 1}\,dx \\ &= \arctan(x - 3) + C\end{aligned} ∫ x 2 − 6 x + 10 1 d x = ∫ ( x − 3 ) 2 + 1 1 d x = arctan ( x − 3 ) + C
Find the indefinite integral.
∫ x 2 − 1 d x \int \sqrt{x^{2} - 1}\,dx ∫ x 2 − 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Integrating by parts makes the original integral I I I
Answer ∫ x 2 − 1 d x = 1 2 ( x x 2 − 1 − log ∣ x + x 2 − 1 ∣ ) + C \int \sqrt{x^{2} - 1}\,dx = \frac{1}{2}(x\sqrt{x^{2} - 1} - \log|x + \sqrt{x^{2} - 1}|) + C ∫ x 2 − 1 d x = 2 1 ( x x 2 − 1 − log ∣ x + x 2 − 1 ∣ ) + C (C C C
Solution Let I = ∫ x 2 − 1 d x I = \displaystyle\int \sqrt{x^{2} - 1}\,dx I = ∫ x 2 − 1 d x ( x ) ′ = 1 (x)' = 1 ( x ) ′ = 1 x 2 = ( x 2 − 1 ) + 1 x^2 = (x^{2} - 1) + 1 x 2 = ( x 2 − 1 ) + 1
I = ∫ x 2 − 1 d x = x x 2 − 1 − ∫ x 2 x 2 − 1 d x = x x 2 − 1 − ∫ ( x 2 − 1 + 1 x 2 − 1 ) d x = x x 2 − 1 − ∫ x 2 − 1 d x − ∫ 1 x 2 − 1 d x \begin{aligned}I &= \int \sqrt{x^{2} - 1}\,dx \\ &= x\sqrt{x^{2} - 1} - \int \frac{x^{2}}{\sqrt{x^{2} - 1}}\,dx \\ &= x\sqrt{x^{2} - 1} - \int \left(\sqrt{x^{2} - 1} + \frac{1}{\sqrt{x^{2} - 1}}\right)\,dx \\ &= x\sqrt{x^{2} - 1} - \int \sqrt{x^{2} - 1}\,dx - \int \frac{1}{\sqrt{x^{2} - 1}}\,dx\end{aligned} I = ∫ x 2 − 1 d x = x x 2 − 1 − ∫ x 2 − 1 x 2 d x = x x 2 − 1 − ∫ ( x 2 − 1 + x 2 − 1 1 ) d x = x x 2 − 1 − ∫ x 2 − 1 d x − ∫ x 2 − 1 1 d x For the last integral, by the formula ∫ d x x 2 + A = log ∣ x + x 2 + A ∣ + C \displaystyle\int \frac{dx}{\sqrt{x^2 + A}} = \log\left|x + \sqrt{x^2 + A}\right| + C ∫ x 2 + A d x = log x + x 2 + A + C
2 I = x x 2 − 1 − log ∣ x + x 2 − 1 ∣ + C 1 2I = x\sqrt{x^{2} - 1} - \log|x + \sqrt{x^{2} - 1}| + C_1 2 I = x x 2 − 1 − log ∣ x + x 2 − 1 ∣ + C 1 Therefore (C C C
I = 1 2 ( x x 2 − 1 − log ∣ x + x 2 − 1 ∣ ) + C I = \frac{1}{2}(x\sqrt{x^{2} - 1} - \log|x + \sqrt{x^{2} - 1}|) + C I = 2 1 ( x x 2 − 1 − log ∣ x + x 2 − 1 ∣ ) + C Practice problems
Find the indefinite integral.
∫ 2 x 2 + 4 d x \int \frac{2}{x^{2} + 4}\,dx ∫ x 2 + 4 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint An antiderivative of 1 x 2 + a 2 \frac{1}{x^2 + a^2} x 2 + a 2 1 1 a arctan x a \frac{1}{a}\arctan\frac{x}{a} a 1 arctan a x x = a tan θ x = a\tan\theta x = a tan θ
Answer ∫ 2 x 2 + 4 d x = arctan x 2 + C \int \frac{2}{x^{2} + 4}\,dx = \arctan\frac{x}{2} + C ∫ x 2 + 4 2 d x = arctan 2 x + C (C C C
Solution Use the formula ∫ d x x 2 + a 2 = 1 a arctan x a + C \displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C ∫ x 2 + a 2 d x = a 1 arctan a x + C a > 0 a > 0 a > 0
∫ 2 x 2 + 4 d x = arctan x 2 + C \begin{aligned}\int \frac{2}{x^{2} + 4}\,dx &= \arctan\frac{x}{2} + C\end{aligned} ∫ x 2 + 4 2 d x = arctan 2 x + C
Find the indefinite integral.
∫ 1 − x 2 + 4 x d x \int \frac{1}{\sqrt{-x^{2} + 4x}}\,dx ∫ − x 2 + 4 x 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Complete the square under the root to get the form a 2 − ( x − p ) 2 a^2 - (x - p)^2 a 2 − ( x − p ) 2
Answer ∫ 1 − x 2 + 4 x d x = arcsin x − 2 2 + C \int \frac{1}{\sqrt{-x^{2} + 4x}}\,dx = \arcsin\frac{x - 2}{2} + C ∫ − x 2 + 4 x 1 d x = arcsin 2 x − 2 + C (C C C
Solution Completing the square under the root, − x 2 + 4 x = 4 − ( x − 2 ) 2 -x^{2} + 4x = 4 - (x - 2)^{2} − x 2 + 4 x = 4 − ( x − 2 ) 2 ∫ d x a 2 − x 2 = arcsin x a + C \displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C ∫ a 2 − x 2 d x = arcsin a x + C a > 0 a > 0 a > 0 x − 2 x - 2 x − 2 x x x
∫ 1 − x 2 + 4 x d x = ∫ 1 4 − ( x − 2 ) 2 d x = arcsin x − 2 2 + C \begin{aligned}\int \frac{1}{\sqrt{-x^{2} + 4x}}\,dx &= \int \frac{1}{\sqrt{4 - (x - 2)^{2}}}\,dx \\ &= \arcsin\frac{x - 2}{2} + C\end{aligned} ∫ − x 2 + 4 x 1 d x = ∫ 4 − ( x − 2 ) 2 1 d x = arcsin 2 x − 2 + C
Find the indefinite integral.
∫ 1 ( x 2 + 1 ) ( x 2 + 4 ) d x \int \frac{1}{(x^{2} + 1)(x^{2} + 4)}\,dx ∫ ( x 2 + 1 ) ( x 2 + 4 ) 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Treat x 2 x^2 x 2
Answer ∫ 1 ( x 2 + 1 ) ( x 2 + 4 ) d x = 1 3 ( arctan x − 1 2 arctan x 2 ) + C \int \frac{1}{(x^{2} + 1)(x^{2} + 4)}\,dx = \frac{1}{3}\left(\arctan x - \frac{1}{2}\arctan\frac{x}{2}\right) + C ∫ ( x 2 + 1 ) ( x 2 + 4 ) 1 d x = 3 1 ( arctan x − 2 1 arctan 2 x ) + C (C C C
Solution Treating x 2 x^2 x 2
1 ( x 2 + 1 ) ( x 2 + 4 ) = 1 3 ( 1 x 2 + 1 − 1 x 2 + 4 ) \begin{aligned}\frac{1}{(x^{2} + 1)(x^{2} + 4)} &= \frac{1}{3}\left(\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}\right)\end{aligned} ( x 2 + 1 ) ( x 2 + 4 ) 1 = 3 1 ( x 2 + 1 1 − x 2 + 4 1 ) Using the formula ∫ d x x 2 + a 2 = 1 a arctan x a + C \displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C ∫ x 2 + a 2 d x = a 1 arctan a x + C a > 0 a > 0 a > 0
∫ 1 ( x 2 + 1 ) ( x 2 + 4 ) d x = ∫ 1 3 ( 1 x 2 + 1 − 1 x 2 + 4 ) d x = 1 3 ( arctan x − 1 2 arctan x 2 ) + C \begin{aligned}\int \frac{1}{(x^{2} + 1)(x^{2} + 4)}\,dx &= \int \frac{1}{3}\left(\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}\right)\,dx \\ &= \frac{1}{3}\left(\arctan x - \frac{1}{2}\arctan\frac{x}{2}\right) + C\end{aligned} ∫ ( x 2 + 1 ) ( x 2 + 4 ) 1 d x = ∫ 3 1 ( x 2 + 1 1 − x 2 + 4 1 ) d x = 3 1 ( arctan x − 2 1 arctan 2 x ) + C