Method of Substitutions for Integration : \(t=tan(x)\) substitution

\[\int \frac{1}{1+sin^2(x)}dx = \frac{1}{\sqrt{2}}tan^{-1} \begin{pmatrix} \sqrt{2}tan(x) \end{pmatrix} + C\]

We start by writing the integral in terms of \(t\).

Using the right-angled triangle, shown here, we can state: \[sin(x) = \frac{t}{\sqrt{1+t^2}}\] and therefore: \[sin^2(x) = \frac{t^2}{1+t^2}\] Furthermore, since \(t = tan(x)\) we have \(\frac{dt}{dx} = 1 +tan^2(x)\), that's: \(\frac{dt}{dx} = 1 + t^2\) and therefore: \[dx = \frac{1}{1+t^2}dt\] Using those results we can now write the integral in terms of \(t\): \[\begin{aligned} \int \frac{1}{1+sin^2(x)}dx & = \int \frac{1}{1 + \frac{t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{\frac{1+t^2}{1+t^2} + \frac{t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{\frac{1+2t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1+t^2}{1+2t^2}.\frac{1}{1+t^2}dt \\ \int \frac{1}{1+sin^2(x)}dx & = \int \frac{1}{1+2t^2}dt \end{aligned}\]

Now that the integral is written in terms of \(t\), we integrate with respect to \(t\): \[\begin{aligned} \int \frac{1}{1+sin^2(x)}dx & = \int \frac{1}{1+2t^2}dt \\ & = \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{1+\begin{pmatrix} \sqrt{2}t \end{pmatrix}^2}dt \\ \int \frac{1}{1+sin^2(x)}dx & = \frac{1}{\sqrt{2}}tan^{-1} \begin{pmatrix} \sqrt{2}t \end{pmatrix} + C \end{aligned}\] Finally, we use the fact that \(t = tan(x)\) to write our answer in terms of \(x\): \[\int \frac{1}{1+sin^2(x)}dx = \frac{1}{\sqrt{2}}tan^{-1} \begin{pmatrix} \sqrt{2}tan(x) \end{pmatrix} + C\]

\[\int \frac{1}{1+cos^2(x)}dx = \frac{\sqrt{2}}{2} .tan^{-1}\begin{pmatrix} \frac{tan(x)}{\sqrt{2}}\end{pmatrix} + C \]

We start by writing the integral in terms of \(t\).

Using the right-angled triangle, shown here, we can state: \[cos(x) = \frac{1}{\sqrt{1+t^2}}\] and therefore: \[cos^2(x) = \frac{1}{1+t^2}\] Furthermore, since \(t = tan(x)\) we have \(\frac{dt}{dx} = 1 +tan^2(x)\), that's: \(\frac{dt}{dx} = 1 + t^2\) and therefore: \[dx = \frac{1}{1+t^2}dt\] Using those results we can now write the integral in terms of \(t\): \[\begin{aligned} \int \frac{1}{1+cos^2(x)}dx & = \int \frac{1}{1 + \frac{1}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{\frac{1+t^2}{1+t^2} + \frac{1}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{\frac{2 + t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1+t^2}{2 + t^2}.\frac{1}{1+t^2}dt \\ \int \frac{1}{1+cos^2(x)}dx & = \int \frac{1}{2 + t^2}dt \end{aligned}\] We now integrate with respect to \(t\): \[\begin{aligned} \int \frac{1}{1+cos^2(x)}dx & = \int \frac{1}{2 + t^2}dt \\ & = \frac{1}{2}\int \frac{1}{1 + \begin{pmatrix}\frac{t}{\sqrt{2}} \end{pmatrix}^2}dt \\ & = \frac{1}{2} \times \sqrt{2}.tan^{-1}\begin{pmatrix} \frac{t}{\sqrt{2}}\end{pmatrix} + C \\ \int \frac{1}{1+cos^2(x)}dx & = \frac{\sqrt{2}}{2} .tan^{-1}\begin{pmatrix} \frac{t}{\sqrt{2}}\end{pmatrix} + C \\ \end{aligned}\] Finally, using the fact that \(t = tan(x)\), we write the answer in terms of \(x\): \[\int \frac{1}{1+cos^2(x)}dx = \frac{\sqrt{2}}{2} .tan^{-1}\begin{pmatrix} \frac{tan(x)}{\sqrt{2}}\end{pmatrix} + C \]

\[\int \frac{1}{3cos^2(x)+sin^2(x)}dx = \frac{1}{\sqrt{3}}tan^{-1}\begin{pmatrix} \frac{tan(x)}{\sqrt{3}} \end{pmatrix} + C \]

We start by writing the integral in terms of \(t\).

Using the right-angled triangle, shown here, we can state: \[cos(x) = \frac{1}{\sqrt{1+t^2}}\] and therefore: \[cos^2(x) = \frac{1}{1+t^2}\] In a similar way, we can also state: \[sin(x) = \frac{t}{\sqrt{1+t^2}}\] and therefore: \[sin^2(x) = \frac{t^2}{1+t^2}\] Furthermore, since \(t = tan(x)\) we have \(\frac{dt}{dx} = 1 +tan^2(x)\), that's: \(\frac{dt}{dx} = 1 + t^2\) and therefore: \[dx = \frac{1}{1+t^2}dt\] Using those results we can now write the integral in terms of \(t\): \[\begin{aligned} \int \frac{1}{3cos^2(x)+sin^2(x)}dx & = \int \frac{1}{ \frac{3}{1+t^2} + \frac{t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{ \frac{3 + t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1+t^2}{ 3 + t^2}.\frac{1}{1+t^2}dt \\ \int \frac{1}{3cos^2(x)+sin^2(x)}dx & = \int \frac{1}{3 + t^2}dt \end{aligned}\] We now integrate with respect to \(t\): \[\begin{aligned} \int \frac{1}{3cos^2(x)+sin^2(x)}dx & = \int \frac{1}{3 + t^2}dt \\ & = \int \frac{1}{\begin{pmatrix} \sqrt{3} \end{pmatrix}^2 + t^2} dt \\ \int \frac{1}{3cos^2(x)+sin^2(x)}dx & = \frac{1}{\sqrt{3}}tan^{-1}\begin{pmatrix} \frac{t}{\sqrt{3}} \end{pmatrix} + C \end{aligned}\] Finally, using the fact that \(t = tan(x)\), we write the answer in terms of \(x\): \[\int \frac{1}{3cos^2(x)+sin^2(x)}dx = \frac{1}{\sqrt{3}}tan^{-1}\begin{pmatrix} \frac{tan(x)}{\sqrt{3}} \end{pmatrix} + C \]

\[\int \frac{dx}{3cos^2(x)+sin^2(x)} = \frac{1}{\sqrt{3}}tan^{-1}\begin{pmatrix} \frac{tan(x)}{\sqrt{3}} \end{pmatrix} + C \]

\[\int \frac{dx}{2sin^2(x) + cos^2(x) + 2} = \frac{1}{2\sqrt{3}} .tan^{-1} \begin{pmatrix} \frac{2tan(x)}{\sqrt{3}} \end{pmatrix} + C\]

We start by writing the integral in terms of \(t\).

Using the right-angled triangle, shown here, we can state: \[cos(x) = \frac{1}{\sqrt{1+t^2}}\] and therefore: \[cos^2(x) = \frac{1}{1+t^2}\] In a similar way, we can also state: \[sin(x) = \frac{t}{\sqrt{1+t^2}}\] and therefore: \[sin^2(x) = \frac{t^2}{1+t^2}\] Furthermore, since \(t = tan(x)\) we have \(\frac{dt}{dx} = 1 +tan^2(x)\), that's: \(\frac{dt}{dx} = 1 + t^2\) and therefore: \[dx = \frac{1}{1+t^2}dt\] Using those results we can now write the integral in terms of \(t\): \[\begin{aligned} \int \frac{dx}{2sin^2(x) + cos^2(x) + 2} & = \int \frac{1}{\frac{2t^2}{1+t^2}+\frac{1}{1+t^2} + 2}.\frac{1}{1+t^2} dt \\ & = \int \frac{1}{\frac{2t^2}{1+t^2} + \frac{1}{1+t^2} + \frac{2 + 2t^2}{1 + t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1}{\frac{3 + 4t^2}{1+t^2}}.\frac{1}{1+t^2}dt \\ & = \int \frac{1 + t^2}{3 + 4t^2}.\frac{1}{1+t^2}dt \\ \int \frac{dx}{2sin^2(x) + cos^2(x) + 2} & = \int \frac{1 }{3 + 4t^2}.dt \end{aligned}\] We now integrate with respect to \(t\): \[\begin{aligned} \int \frac{dx}{2sin^2(x) + cos^2(x) + 2} & = \int \frac{1 }{3 + 4t^2}.dt \\ & = \frac{1}{3}\int \frac{1}{1 + \frac{4}{3}t^2}dt \\ & = \frac{1}{3}\int \frac{1}{1 + \begin{pmatrix}\sqrt{\frac{4}{3}}t\end{pmatrix}^2}dt \\ & = \frac{1}{3} \times \sqrt{\frac{3}{4}}.tan^{-1} \begin{pmatrix} \sqrt{\frac{4}{3}} t \end{pmatrix} + C \\ \int \frac{dx}{2sin^2(x) + cos^2(x) + 2} & = \frac{1}{2\sqrt{3}} .tan^{-1} \begin{pmatrix} \frac{2t}{\sqrt{3}} \end{pmatrix} + C \end{aligned}\] Finally, using the fact that \(t = tan(x)\), we write the answer in terms of \(x\): \[\int \frac{dx}{2sin^2(x) + cos^2(x) + 2} = \frac{1}{2\sqrt{3}} .tan^{-1} \begin{pmatrix} \frac{2tan(x)}{\sqrt{3}} \end{pmatrix} + C\]