Question
Question: \[\int {\dfrac{1}{{\sin x\sqrt {\sin x\cos x} }}} dx = \] A) \[\sqrt {\tan x} + c\] B) \[2\sqrt ...
∫sinxsinxcosx1dx=
A) tanx+c
B) 2tanx+c
C) −2cotx+c
D) 2cotx+c
Solution
The given function is indefinite since there is no limit given. Indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.
Complete step by step answer:
Let the given integral be I such that:
I=∫sinxsinxcosx1dx
Multiply numerator and the denominator of the function with sec2x, hence we get
I=∫sinxsinxcosx×sec2xsec2xdx
Now this can be written as
I=∫tanxtanxsec2xdx
Which is equivalent to
I=∫tan23xsec2xdx
Now let
So by differentiating we get
⇒tanx=t2 ⇒sec2x=2tdxdt ⇒dx=sec2x2tdtNow substitute the value of sec2xdx=2tdtin the integral function, hence we get
I=∫t3sec2x×sec2x2tdt =∫t22dtBy integrating, we get
I=∫t22dt =−t2+cNow substitute the value oft=tanx, we get
I=−t2+c =−tanx2+cThis can be written as
I=−tanx2+c =−2cotx+cHence
⇒ ∫sinxsinxcosx1dx=−2cotx+c. Hence Option 3 is correct.
Note: While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.