If $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$ = $A\sqrt{7-6x-x^2} + Bsin^{-1} \left( \frac{x+3}{4} \right) + C$ (Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
Integral Calculation: $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$
The goal is to find the coefficients $A$ and $B$ in the expression:
$ \int \frac{2x+5}{\sqrt{7-6x-x^2}} dx = A\sqrt{7-6x-x^2} + B\sin^{-1} \left( \frac{x+3}{4} \right) + C $
Step 1: Complete the Square
First, simplify the quadratic expression under the square root:
$ 7-6x-x^2 = -(x^2+6x-7) $
Complete the square for $x^2+6x$: $(x+3)^2 = x^2+6x+9$. So, $x^2+6x = (x+3)^2 - 9$.
$ -( (x+3)^2 - 9 - 7 ) = -( (x+3)^2 - 16 ) = 16 - (x+3)^2 $
The integral becomes:
$ \int \frac{2x+5}{\sqrt{16-(x+3)^2}} dx $
Step 2: Manipulate the Numerator
Express the numerator $2x+5$ in terms of $(x+3)$ to relate it to the derivative of the expression inside the square root, or to simplify using substitution.
Let $y = x+3$. Then $x = y-3$.
$ 2x+5 = 2(y-3)+5 = 2y-6+5 = 2y-1 $
Substitute back $y = x+3$:
$ 2x+5 = 2(x+3) - 1 $
Step 3: Split the Integral
Rewrite the integral using the manipulated numerator:
$ \int \frac{2(x+3)-1}{\sqrt{16-(x+3)^2}} dx = \int \frac{2(x+3)}{\sqrt{16-(x+3)^2}} dx - \int \frac{1}{\sqrt{16-(x+3)^2}} dx $
Step 4: Evaluate the First Integral
Consider the first part: $\int \frac{2(x+3)}{\sqrt{16-(x+3)^2}} dx$.
Let $u = 16-(x+3)^2$. Then $du = -2(x+3)dx$. This implies $2(x+3)dx = -du$.
The integral becomes:
$ \int \frac{-du}{\sqrt{u}} = - \int u^{-1/2} du = - \frac{u^{1/2}}{1/2} + C_1 = -2\sqrt{u} + C_1 $
Substitute back $u = 16-(x+3)^2 = 7-6x-x^2$:
$ -2\sqrt{7-6x-x^2} + C_1 $
Comparing this with $A\sqrt{7-6x-x^2}$, we find $A = -2$.
Step 5: Evaluate the Second Integral
Consider the second part: $- \int \frac{1}{\sqrt{16-(x+3)^2}} dx$.
This integral is a standard form: $\int \frac{1}{\sqrt{a^2-y^2}} dy = \sin^{-1}\left(\frac{y}{a}\right)$.
Here, $y = x+3$ and $a^2=16$, so $a=4$.
$ - \int \frac{1}{\sqrt{16-(x+3)^2}} dx = - \sin^{-1}\left(\frac{x+3}{4}\right) + C_2 $
Comparing this with $B\sin^{-1} \left( \frac{x+3}{4} \right)$, we find $B = -1$.
Step 6: Combine Results and Final Answer
Combining both parts, the integral is:
$ -2\sqrt{7-6x-x^2} - \sin^{-1}\left(\frac{x+3}{4}\right) + C $
Comparing with the given form $A\sqrt{7-6x-x^2} + B\sin^{-1} \left( \frac{x+3}{4} \right) + C$, we have:
$ A = -2 $
$ B = -1 $
Therefore, the ordered pair $(A, B)$ is $(-2, -1)$.
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