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Example 2: Trigonometric substitution

For the integral ∫ 0 1 1 − x 2 d x, {\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}\,dx,} a variation of the above procedure is needed. The substitution x = sin u {\displaystyle x=\sin u} implying d x = cos u d u {\displaystyle dx=\cos u\,du} is useful because 1 − sin 2 u = cos u. {\textstyle {\sqrt {1-\sin ^{2}u}}=\cos u.} We thus have: ∫ 0 1 1 − x 2 d x = ∫ 0 π / 2 1 − sin 2 u cos u d u = ∫ 0 π / 2 cos 2 u d u = 0 π / 2 = π 4 + 0 = π 4. {\displaystyle {\begin{aligned}\int _{0}^{1}{\sqrt {1-x^{2}}}\ dx&=\int _{0}^{\pi /2}{\sqrt {1-\sin ^{2}u}}\cos u\ du\\&=\int _{0}^{\pi /2}\cos ^{2}u\ du\\&=\left_{0}^{\pi /2}\\&={\frac {\pi }{4}}+0\\&={\frac {\pi }{4}}.\end{aligned}}}

The resulting integral can be computed using integration by parts or a double angle formula, 2 cos 2 u = 1 + cos (2 u), {\textstyle 2\cos ^{2}u=1+\cos(2u),} followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or π 4. {\displaystyle {\tfrac {\pi }{4}}.}