Question
Evaluvate the following intregals:
$\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$
$=\int\frac{1}{\text{P}+\frac{q\sin\text{x}}{\cos\text{x}}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\ \text{dx}$
Let $\cos\text{x}=\text{A}(q\sin\text{x}+\text{p}\cos\text{x})+\text{B}(q\cos\text{x}-\text{p}\sin\text{x})$
$\Rightarrow\cos\text{x}=(\text{Ap}+\text{Bq})\cos\text{x}+(\text{Aq}-\text{Bp})\sin\text{x}$
Compairing coefficient of like terms
$\text{Ap}+\text{Bq}=0\dots(1)$
$\text{Aq}+\text{Bp}=1\dots(2)$
Multipiying eq (1) by p and eq (2) by q and then adding
$\Rightarrow\text{Ap}^2+\text{Bpq}=\text{p}$
$\Rightarrow\text{Aq}^2+\text{Bpq}=0$
$\Rightarrow\text{A}=\frac{\text{p}}{\text{p}^2\text{q}^2}$
Putting value of A in eq (1)
$\frac{\text{p}^2}{\text{p}^2+\text{q}^2}+\text{Bq}=1$
$\Rightarrow\text{Bq}=1-\frac{\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{Bq}=\frac{\text{p}^2+\text{q}^2-\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{B}=\frac{\text{q}}{\text{p}^2+\text{q}^2}$
$\therefore\text{I}=\int\Big[\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\times\frac{(\text{q}\sin\text{x}+\text{p}\cos\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\times\frac{(q\cos\text{x}-\text{p}\sin\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}\Big]\text{dx}$
$=\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\text{dx}+\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\int\Big(\frac{q\cos\text{x}-\text{p}\sin\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\Big)\text{dx}$
Putting $\text{q}\sin\text{x}+\text{p}\cos\text{x}=\text{t}$
$\Rightarrow(\text{q}\cos\text{x}-\text{p}\sin\text{x})\text{ dx}=\text{dt}$
$\therefore\text{I}=\frac{\text{p}}{\text{q}+\text{q}^2}\int\text{dx}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{\text{p}}{\text{q}+\text{q}^2}\text{x}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\ln|\text{q}\sin\text{x}+\text{p}\cos\text{x}|+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
Let R be a relation on the set A of ordered pair of integers defined by (x, y)R(u, v) if xv = yu. Show that R is an equivalence relation.
If a, b and c are real numbers, and
$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,$
Show that either a + b + c = 0 or a = b = c.
If the straight line $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$ touches the curve $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1,$ then prove that $\text{a}\cos^2\alpha-\text{b}^2\sin^2\alpha=\text{p}^2.$
Using differentials, find the approximate values of the following:
$\sqrt{0.48}$
find the area of the region bound by the curve $x = at^2, y = 2$ at between the ordinatrs corresponding $t = 1$ and $t = 2$.
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients-calcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
 
Food I
(per Ib)
Food II
(per Ib)
Minimum daliy requarement
for the nutrient
Calcium
10
5
20
Protein
5
4
20
Calories
2
6
13
Price (Rs)
60
100
 
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
Find the area common to the circle $x^2 - y^2 = 16 a^2$ and the parabola $y^2 = 6x$.