To integrate by parts, we observe the derivative of the denominator term x sin x + cos x. Its derivative is (d)/(dx)(x sin x + cos x) = x cos x + sin x - sin x = x cos x. Thus, we rewrite the numerator x² as (x)/(cos x) · (x cos x) so that one part is directly integrable.

To integrate by parts, we observe the derivative of the denominator term x sin x + cos x. Its derivative is (d)/(dx)(x sin x + cos x) = x cos x + sin x - sin x = x cos x. Thus, we rewrite the numerator x² as (x)/(cos x) · (x cos x) so that one part is directly integrable.

Taking f(x) = (x)/(cos x) = x sec x as the first function and g(x) = (x cos x)/((x sin x + cos x)²) as the second function, we apply the integration by parts rule: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [ (d)/(dx)f(x) ∫ g(x) dx ] dx.

Taking f(x) = (x)/(cos x) = x sec x as the first function and g(x) = (x cos x)/((x sin x + cos x)²) as the second function, we apply the integration by parts rule: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [ (d)/(dx)f(x) ∫ g(x) dx ] dx.

Differentiating f(x) = (x)/(cos x) using the quotient rule gives (1 · cos x - x(-sin x))/(cos² x) = (cos x + x sin x)/(cos² x). Substituting this into the remaining integral leads to an exact cancellation of the factor (x sin x + cos x).

Differentiating f(x) = (x)/(cos x) using the quotient rule gives (1 · cos x - x(-sin x))/(cos² x) = (cos x + x sin x)/(cos² x). Substituting this into the remaining integral leads to an exact cancellation of the factor (x sin x + cos x).

The remaining integral simplifies to ∫ (1)/(cos² x) dx = ∫ sec² x dx = tan x. Adding the constant of integration C, we obtain the final answer.

$\int \frac{x^2}{(x \sin x + \cos x)^2} \, dx$

Answer: $\frac{-x}{x \sin x \cos x + \cos^2 x} + \tan x + C = -\frac{x \sec x}{x \sin x + \cos x} + \tan x + C$

Step-by-step solution

Step 1: Rewrite integrand by multiplying and dividing by cosine

To integrate by parts, we observe the derivative of the denominator term $x \sin x + \cos x$ . Its derivative is $\frac{d}{dx}(x \sin x + \cos x) = x \cos x + \sin x - \sin x = x \cos x$ . Thus, we rewrite the numerator $x^2$ as $\frac{x}{\cos x} \cdot (x \cos x)$ so that one part is directly integrable.

Step 2: Find the integral of the second function

Let $t = x \sin x + \cos x$ . Then $dt = (x \cos x + \sin x - \sin x) \, dx = x \cos x \, dx$ . The integral becomes $\int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{x \sin x + \cos x}$ .

Step 3: Apply integration by parts

Taking $f(x) = \frac{x}{\cos x} = x \sec x$ as the first function and $g(x) = \frac{x \cos x}{(x \sin x + \cos x)^2}$ as the second function, we apply the integration by parts rule: $\int f(x) g(x) \, dx = f(x) \int g(x) \, dx - \int \left[ \frac{d}{dx}f(x) \int g(x) \, dx \right] dx$ .

Step 4: Differentiate the first function and simplify the remaining integral

Differentiating $f(x) = \frac{x}{\cos x}$ using the quotient rule gives $\frac{1 \cdot \cos x - x(-\sin x)}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}$ . Substituting this into the remaining integral leads to an exact cancellation of the factor $(x \sin x + \cos x)$ .

Step 5: Evaluate the remaining integral and conclude

The remaining integral simplifies to $\int \frac{1}{\cos^2 x} \, dx = \int \sec^2 x \, dx = \tan x$ . Adding the constant of integration $C$ , we obtain the final answer.

Solve your own maths question free →

∫x2(xsin⁡x+cos⁡x)2 dx\int \frac{x^2}{(x \sin x + \cos x)^2} \, dx∫(xsinx+cosx)2x2​dx