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I am high school student and found a question in NCERT Class 12 book.

The question is:

$$\int\frac{1}{1-\tan(x)}dx$$

After simplifying the question, I got:

$$\frac{1}{2}\left(\int\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}+1\right)dx$$

I then used substitution method.

$$\cos(x)-\sin(x)=t$$

$$dx=\frac{-1}{\cos(x)+\sin(x)}\cdot dt$$

After solving, I got

$$\frac{1}{2}(\cos(x)-\sin(x))-\frac{1}{2}(\ln(|\cos(x)-\sin(x|))+C$$

But the answer in the book is different:

$$\frac{x}{2}-\frac{1}{2}\ln(|\cos(x)-\sin(x)|+C$$

I think that the textbook answer is different because they integrated 1 with respect to x whereas I integrated 1 with respect to t. What's the right way to integrate?

Random Person
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    The book is right. $\int 1 dx=x+C$. If you want to use integration w.r.t $t$ you should write $\int 1 \frac {dx} {dt}dt$ but that makes it complicated. – Kavi Rama Murthy May 31 '22 at 09:03

1 Answers1

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After simplifying the question, I got: $$\frac{1}{2}\left(\int\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}+1\right)dx $$ I then used substitution method. $$\cos(x)-\sin(x)=t$$

Note that $$\frac{1}{2}\int\left(\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}+1\right)\,\mathrm dx = \frac12\int\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}\,\mathrm dx+ \frac12\int1\,\mathrm dx$$ and that $$\frac12\int1\,\mathrm dx=\frac12x\ne\frac12t=\frac12\left(\cos(x)-\sin(x)\right).$$ Note also that $$\frac12\int1\,\mathrm dx =\frac12\int\frac{-1}{\cos(x)+\sin(x)}\,\mathrm dt \ne\frac12t.$$

ryang
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