\(ln|(1-tanx)|+C \)
\(ln |3+sinx|+C\)
\(ln |2+tanx|+C\)
\(ln |1+2secx|+C\)
\(ln |2-tanx|+C\)
Given that
\(∫\dfrac{1}{cosx(sinx+2cosx)} dx\)
take, \(I = ∫\dfrac{1}{cosx(sinx+2cosx)} dx\)
\(= ∫\dfrac{1}{cosx.sinx+2cos^{2}x)} dx\)
\(= ∫\dfrac{sec^{2}x}{tanx+2} dx\) ⇢⇢[by dividing \(cos^{2}x\) on both numerator and denominator]
take \(tanx +2 = u\), then derivate both sides w.r.t x we get
\(sec^{2}x dx= du\)
Bye applying this we can write , \(∫\dfrac{1}{u} du\)
\(= ln|u|\)
\(= ln|(tanx+2)|+C\)
\(= ln|(2+tanx)|+C\)
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C