Trigonometric substitution
In mathematics, a trigonometric substitution replaces a trigonometric function for another expression.
In calculus, trigonometric substitutions are a technique for evaluating integrals.
In this case, an expression involving a radical function is replaced with a trigonometric one.
Trigonometric identities may help simplify the answer.
[1][2] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.
by using the inverse sine function.
For a definite integral, one must figure out how the bounds of integration change.
Some care is needed when picking the bounds.
Neglecting this restriction, one might have picked
which would have resulted in the negative of the actual value.
Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions.
For a definite integral, the bounds change once the substitution is performed and are determined using the equation
Alternatively, apply the boundary terms directly to the formula for the antiderivative.
On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields
For a definite integral, the bounds change once the substitution is performed and are determined using the equation
Alternatively, apply the boundary terms directly to the formula for the antiderivative.
Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields
The integral of secant cubed may be evaluated using integration by parts.
( sec θ tan θ + ln
can also be evaluated by partial fractions rather than trigonometric substitutions.
One may evaluate the integral of the secant function by multiplying the numerator and denominator by
and the integral of secant cubed by parts.
( sec θ tan θ + ln
( sec θ tan θ − ln
Substitution can be used to remove trigonometric functions.
{\displaystyle {\begin{aligned}\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du&&u=\sin(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\mp {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du&&u=\cos(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {2}{1+u^{2}}}f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du&&u=\tan \left({\frac {x}{2}}\right)\\[6pt]\end{aligned}}}
The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.
Substitutions of hyperbolic functions can also be used to simplify integrals.
{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}+x^{2}}}}&=\int {\frac {a\cosh u\,du}{\sqrt {a^{2}+a^{2}\sinh ^{2}u}}}\\[6pt]&=\int {\frac {\cosh {u}\,du}{\sqrt {1+\sinh ^{2}{u}}}}\\[6pt]&=\int {\frac {\cosh {u}}{\cosh u}}\,du\\[6pt]&=u+C\\[6pt]&=\sinh ^{-1}{\frac {x}{a}}+C.\end{aligned}}}
If desired, this result may be further transformed using other identities, such as using the relation
