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These identities supply the formulas used in the tangent half-angle substitution. The half-angle identities include

If $F^{\prime }=f$ and $g$ is differentiable, the chain rule gives $(F\circ g{)}^{\prime }(x)=f(g(x))g^{\prime }(x)$. Taking antiderivatives gives the substitution ruleWriting $u=g(x)$ and $du=g^{\prime }(x)dx$ records this chain rule in reverse. For a continuous function $f$ and continuously differentiable $g$, the fundamental theorem of calculus gives the corresponding definite integral identityThe order of the transformed endpoints determines the sign. For example, ${\int }_0^{1}2x\cos (x^2)dx={\int }_0^1\cos udu=\sin 1$. If the substitution is used in the opposite direction by solving $x$ as a function of $u$, one must restrict to a suitable inverse branch; the tangent half-angle substitution illustrates why separate branches can be necessary even when the original integrand is smooth. The general change of variables formula extends substitution to higher dimensions.

The tangent half-angle substitution $u=\tan (t/2)$ maps $0\le t<\pi$ to $0\le u<\infty$, withBecause $a>1$, both $a+1$ and $a-1$ are positive. ThusEvaluating the improper integral givesAlthough the change of variable has an infinite endpoint, the original integrand is continuous on the closed interval, and the transformed integrand decays as $u^{-2}$, so the limiting substitution is justified.

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints $x=0,\pi /2$ become $t=0,1$, respectively. HenceCompleting the square, $t^2+t+1=(t+1/2{)}^2+3/4$, then gives

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints become zero and one, soUse completing the square, $t^2+t+1=(t+1/2{)}^2+3/4$, and integrate:Thus both integrals equal $\boxed{\pi /(3\sqrt{3})}$.

The tangent half-angle substitution gives $\sin \theta =2t/(1+t^2)$ and $\cos \theta =(1-t^2)/(1+t^2)$. Dividing and then taking the reciprocal square of the cosine givesThese formulas apply where $t$ is defined and $t\ne \pm 1$; at $t=\pm 1$ the original tangent and secant function are undefined because $\cos \theta =0$.

Use the tangent half-angle substitution $t=\tan (x/2)$, so $\cos x=(1-t^2)/(1+t^2)$ and $dx=2dt/(1+t^2)$. ThenThe inverse tangent antiderivative givesThis expression applies on each interval $((2j-1)\pi ,(2j+1)\pi )$ for an integer $j$. Although $5+3\cos x\ge 2$ everywhere, the chosen inverse tangent expression jumps at the endpoints because the substitution has a pole there. A global continuous antiderivative is obtained by adding $j\pi /2$ on that interval, using one common $C$, and extending by continuity at the endpoints.

The tangent half-angle substitution $t=\tan (x/2)$ converts rational trigonometric expressions into rational functions: $\sin x=2t/(1+t^2)$, $\cos x=(1-t^2)/(1+t^2)$ and $dx=2dt/(1+t^2)$.