Question

Anti-Derivative ∫(1 + tan²x) dx

Answer 1

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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