# SOLVED: Evaluate the integral âˆ« tan(3x) dx

Integral of sin^3x
Integral of sin^3x

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Evaluate the integral
âˆ« tan(3x) dx

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Evaluate the indefinite integral of sec(3x) tan(3x) dx (Use C for the constant of integration).

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Evaluate the integral.âˆ« 5 sec(4x) dxâˆ« 5 tan(3x+Ï€/3) dxâˆ« 5 dxâˆ« 3 tan(5x) dxâˆ« 5(sec(x) + tan(x))^5 dxâˆ« 5 tan(x) + tan(âˆšx) dx

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Evaluate the integral.$$\int(\tan 3 x+\sec 3 x) d x$$

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$\int\left[(\mathrm{d} \mathrm{x}) /(2 \sin \mathrm{x}+3 \cos \mathrm{x})^{2}\right]=\underline{\mathrm{c}}$(a) $-[1 /(2 \tan x+3)]$ (b) $[\overline{1 /(2 \tan x+3})]$(c) $-[1 /\{2(2 \tan \mathrm{x}+3)\}]$(d) $[1 /\{2(2 \tan \mathrm{x}+3)\}]$

Transcript

to solve the integral tan to the power of 5 of 3x dx. So, first of all put 3x is equal to t, then we have on differentiation 3 times of dx is equal to dt. So, therefore, dx is equal to dt over 3. So, integration of tan to the power of 5 of t dt over 3 that is 1 over 3, it will give tan of t multiplied to tan to the power of 4t dt. So, now, here 1 over 3 integration of tan t, it will give sec square t negative 1 to the whole square dt. Expand the whole square, it will give sec to the power 4t negative 2 times of sec square t positive 1 and dt. So, now further open the bracket 1 over 3 in bracket tan t, it can write sec square sec square, sec square is going to be 1 plus tan square and multiply to sec square t dt and negative 2 over 3 integration of tan t sec square of t dt and plus 1 over 3 integration of tan t dt plus c. So, now put tan t is equal to k on differentiation sec square t dt is equal to dk. So, therefore,…

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