Evaluating ∫ ( tan 3 ⁡ x + tan 4 ⁡ x ) d x using substitution t = tan ⁡ x

Question

Evaluating   ∫  (      tan    3    ⁡  x  +      tan    4    ⁡  x  )  d  x using substitution   t  =  tan  ⁡  x

x713x9o7r · Accepted Answer

Let&#039;s wrap it up. Let   t  =  t  a  n  x, so   d  t  =  s  e      c    2    x  d  x  =  (  1  +  t  a      n    2    x  )  d  x  =  (  1  +      t    2    )  d  x, and   d  x  =                    d        t                    1        +                  t          2                    . So   I  =  ∫                              t          3                +                  t          4                            1        +                  t          2                      d  t  =  ∫      (    t    −                  t                  1          +                      t            2                                +          t      2        −    1    +                  1                  1          +                      t            2                                )    d  t  =                    t        3            3        +                    t        2            2        −  t  −                    l        n        (        1        +                  t          2                )            2        +  t  a      n          −      1        t  +  C  =                    t        a                  n          3                x            3        +                    t        a                  n          2                x            2        −  t  a  n  x  −  l  n      |    s  e  c  x      |    +  x  +  C

Tiffany Page · Accepted Answer

If you feel unconfortable with sec you may proceed from your last integral

I = \int (1 + t) t \sin^{2} (\arctan t) d t

by using the following trigonometric identity

\sin^{2} x = \frac{\tan^{2} x}{1 + \tan^{2} x}, x = \arctan t,

which you can derive from the fundamental identity

\sin^{2} x + \cos^{2} x = 1

to obtain

\sin^{2} (\arctan t) = \frac{t^{2}}{1 + t^{2}} .

Consequently

I = \int (1 + t) t \frac{t^{2}}{1 + t^{2}} d t = \int \frac{t^{3} + t^{4}}{1 + t^{2}} d t,

which is integrable by partial fractions. By long division we compute

\frac{t^{4} + t^{3}}{1 + t^{2}} = t^{2} + t - 1 + \frac{- t + 1}{1 + t^{2}} .

So

I = \int (t^{2} + t - 1) d t + \int \frac{- t + 1}{1 + t^{2}} d t .

Since

\begin{array}{rcl} \int \frac{- t + 1}{1 + t^{2}} d t & = & - \frac{1}{2} \int \frac{2 t}{1 + t^{2}} d t + \int \frac{1}{1 + t^{2}} d t \\ = & - \frac{1}{2} \ln (1 + t^{2}) + \arctan t + C, \end{array}

we thus have

\begin{array}{rcl} I & = & \frac{1}{3} t^{3} + \frac{1}{2} t^{2} - t - \frac{1}{2} \ln (1 + t^{2}) + \arctan t + C \\ = & \frac{1}{3} \tan^{3} x + \frac{1}{2} \tan^{2} x - \tan x - \frac{1}{2} \ln (1 + \tan^{2} x) + x + C . \end{array}

Solve int(tan^3 x+tan^4 x)dx using substitution t=tan x

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