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I have to solve this integral $$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$$ It seems that the most convenient way to operate is doing the substitution $ \tan x = t$. Then after some passages the integral becomes: $$ \int \frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2}\, dx $$

Now I'm trying the coefficient A,B,C,D,E and F such as:

$$\frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2} = \frac{A}{t-1}+\frac{2B(t-1)+C}{t^2+1}+ \frac {d}{dx} \frac{D(t-1)^2+E(t-1)+F}{(t-1)(t^2+1)}$$

Perhaps I made mistakes in this last passage because I find some coefficients that don't lead to the result suggested by the book.

Anne
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3 Answers3

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$$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{\left( t-1\right) {{\left( {{t}^{2}}+1\right) }^{2}}}=-\frac{4}{{{t}^{2}}+1}+\frac{4-4 t}{{{\left( {{t}^{2}}+1\right) }^{2}}}+\frac{1}{t-1}$$

Details: $$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{\left( t-1\right) {{\left( {{t}^{2}}+1\right) }^{2}}}=\frac{B t+C}{{{t}^{2}}+1}+\frac{D t+E}{{{\left( {{t}^{2}}+1\right) }^{2}}}+\frac{A}{t-1}$$

$$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{{{t}^{5}}-{{t}^{4}}+2 {{t}^{3}}-2 {{t}^{2}}+t-1}=\frac{\left( B+A\right) {{t}^{4}}+\left( C-B\right) {{t}^{3}}+\left( D-C+B+2 A\right) {{t}^{2}}+\left( E-D+C-B\right) t-E-C+A}{{{t}^{5}}-{{t}^{4}}+2 {{t}^{3}}-2 {{t}^{2}}+t-1}$$

$$\left( B+A\right) {{t}^{4}}+\left( C-B\right) {{t}^{3}}+\left( D-C+B+2 A\right) {{t}^{2}}+\left( E-D+C-B\right) t-E-C+A={{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1$$ Solve system $$-E-C+A=1\operatorname{,}E-D+C-B=4\operatorname{,}D-C+B+2 A=2\operatorname{,}C-B=-4\operatorname{,}B+A=1$$ We get $$D=-4\operatorname{,}B=0\operatorname{,}E=4\operatorname{,}C=-4\operatorname{,}A=1$$

Aleksas Domarkas
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$$2(\sin x-\cos x)\cos x =\sin2x-2\cos^2x =\sin2x-\cos2x+1=y\text{(say)}$$

Now $1+\sin4x=2-(\cos2x-\sin2x)^2=2-(y-1)^2=?$

Again, $\dfrac{2-(y-1)^2}y=-y+2+\dfrac1y$

Finally $\dfrac1y=\dfrac1{(\sin x-\cos x)\cos x}=\dfrac{\sec^2x}{\tan x-1}$

Should I leave it here ?

lab bhattacharjee
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  • can you show me how to solve this integral just doing the substitution $t=\tan x$ seen that it's the only type of substitution i've learned so far? – Anne Mar 25 '21 at 16:40
  • @Anne, That is required for the last part only, right? Why shall we substitute for $$\int\sin2x\ dx,\int\cos2x\ dx$$ – lab bhattacharjee Mar 25 '21 at 17:53
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Substitute $t=2x+\frac\pi4$ as shown below

\begin{align} &\int \frac{1+\sin 4x}{(\sin x -\cos x) \cos x}\, dx\\ =& \int \frac{2(\sin2x +\cos 2x )^2}{\sin2x-\cos2x-1}dx =-\int \frac{2\sin^2t}{1+\sqrt2\cos t}dx\\ =& \int \left(- 1+\sqrt2\cos t -\frac1{1+\sqrt2\cos t}\right)dt\\ = &-t + \sqrt2\sin t - \ln| \frac{\sqrt2+1+\tan\frac t2}{\sqrt2+1-\tan\frac t2}| +C\\ \end{align}

Quanto
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