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Evaluate the indefinite integral:
\(\mathsf{\displaystyle\int\!(1+tan^2\,x)\,dx}\\\\\\ \mathsf{=\displaystyle\int\!\bigg[1+\left(\frac{sin\,x}{cos\,x}\right)^{\!2}\bigg]dx}\\\\\\
\mathsf{=\displaystyle\int\!\bigg[1+\frac{sin^2\,x}{cos^2\,x}\bigg]dx}\)
Reduce both terms to the same common denominator:
\(\mathsf{=\displaystyle\int\!\bigg[\frac{cos^2\,x}{cos^2\,x}+\frac{sin^2\,x}{cos^2\,x}\bigg]dx}\\\\\\
\mathsf{=\displaystyle\int\!\frac{cos^2\,x+sin^2\,x}{cos^2\,x}\,dx\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\
\mathsf{=\displaystyle\int\! \frac{1}{cos^2\,x}\,dx}\\\\\\
\mathsf{=\displaystyle\int\! sec^2\,x\,dx}\)
and that last one is an immediate integral:
\(\mathsf{\displaystyle\int\! sec^2\,x\,dx=tan\,x+C}\)
Therefore,
\(\mathsf{\displaystyle\int\! (1+tan^2\,x)\,dx=tan\,x+C}\) ✔
I hope this helps. =)
Tags: indefinite integral anti-derivative trigonometric trig function tangent secant sine cosine tan sec sin cos differential integral calculus
