Using the table of integrals from Appendix B in the calculus book 10e, which formula would fit the integral `int(7sec^2(x))/(6tanx(2-3tanx)^2)` . How would you solve it. What are the constants? 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that sec x can be replaced by `1/(cos x)` , hence, the integral would look as it follows, such that:

`int (7(1/(cos^2 x)))/(6tan x(2 - 3tan x)^2) dx`

You should perform the following substitution, such that:

`tan x = t`

`(dx)/(cos^2 x) = dt`

You need to replace tan x by t such that:

`7/6 int (dt)/(t(2-3t)^2)`

You need to evaluate the integral using partial fraction decomposition:

`1/(t(2-3t)^2) = A/t + B/(2 - 3t) + C/((2 - 3t)^2)`

`1 = A(2 - 3t)^2 + Bt(2 - 3t) + Ct`

`1 = 4A - 12At + 9At^2 + 2Bt - 3Bt^2 + Ct`

`1 = t^2(9A - 3B) + t(2B + C - 12A) + 4A`

You need to set equal the coefficients of corresponding powers, such that:

`9A - 3B = 0 => 3A = B`

`2B + C - 12A = 0`

`4A = 1 => A = 1/4 => B = 3/4 => C = 3/2`

`1/(t(2-3t)^2) = 1/(4t) + 3/(4(2 - 3t)) + 3/(2(2 - 3t)^2)`

Integrating both sides:

`int (dt)/(t(2-3t)^2) = int 1/(4t) dt + int 3/(4(2 - 3t)) dt + int 3/(2(2 - 3t)^2) dt`

`int (dt)/(t(2-3t)^2) = (1/4) ln|t| + (3/4)*(-1/3) ln|2 - 3t| - (3/2)*(1/3)*1/(2 - 3t) + c`

`int (7(1/(cos^2 x)))/(6tan x(2 - 3tan x)^2) dx= (7/24) ln|tan x| - (7/24) ln|2 - 3tan x| - (3/7)*1/(2 - 3tan x) + C`

Hence, evaluating the given integral, yields `int (7sec^2 x)/(6tan x(2 - 3tan x)^2) dx = (7/24) ln|tan x| - (7/24) ln|2 - 3tan x| - (3/7)*1/(2 - 3tan x) + C.`

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