Q.Find the integral $\int \frac{2-3\sin x}{\cos^2 x}\,dx$

The key insight is that the expression $\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x$ looks like a disguised perfect cube. Recall the identity for the sum of cubes:

But here we have something even more direct: if we set $a = \sin^2 x$ and $b = \cos^2 x$, then $a+b = 1$. The expression $\sin^6 x + \cos^6 x$ is $(\sin^2 x)^3 + (\cos^2 x)^3$. Expanding the cube of a sum gives a neat simplification.

1. Rewrite the integrand using $a = \sin^2 x$, $b = \cos^2 x$. Then $a + b = 1$. The integrand becomes:

2. Recall the identity for $(a+b)^3$:

Since $a+b = 1$, this simplifies to:

That is exactly our integrand!

3. Therefore, the integrand equals 1 for all $x$. So the integral is:

4. Check the options: (A) $-\frac{3}{2}\cos 2x + C$ — no. …

The key here is that the integrand $\frac{1}{(x+1)(x+2)}$ is a proper rational function — the denominator is already factored into two distinct linear factors. When you see this structure, the natural move is partial fraction decomposition, which splits the fraction into a sum of simpler fractions that you can integrate term-by-term using the standard formula $\int \frac{1}{x-a}\,dx = \log|x-a| + c$.

Let’s walk through it.

1. Set up the partial fractions. We want constants $A$ and $B$ such that

Multiply both sides by $(x+1)(x+2)$ to clear denominators:

2. Solve for $A$ and $B$. Expand the right-hand side:

For this to hold for all $x$, the coefficients of $x$ and the constant term must match on both sides. That gives the system:

From the first equation, $B = -A$. Substitute into the second: $2A - A = 1$, so $A = 1$. Then $B = -1$.

3. Rewrite the integral. Now we have

4. Integrate term by term. Each term is a standard logarithmic integral: $$$$

Concept & Intuition

When you see a squared tangent (or secant) of a linear expression, the identity $\tan^2\theta = \sec^2\theta - 1$ is your best friend. Why? Because $\sec^2\theta$ has a simple antiderivative ($\tan\theta$), and the constant $-1$ integrates to a linear term. The inner function $5 - \frac{x}{2}$ is linear, so a simple substitution (or just adjusting for the chain rule) handles the factor.

Step‑by‑step solution

1. Set up the substitution Let $u = 5 - \frac{x}{2}$. Then $du = -\frac{1}{2}\,dx$, so $dx = -2\,du$. The integral becomes

2. Use the Pythagorean identity Recall $\tan^2 u = \sec^2 u - 1$. Substitute:

3. Integrate term by term
   • $\int \sec^2 u\,du = \tan u + C$ (this is a standard derivative fact).
   • $\int 1\,du = u + C$. So we have

4. Substitute back Replace $u$ with $5 - \frac{x}{2}$:

Simplify the linear part: $2\cdot5 = 10$ and $2\cdot\left(-\frac{x}{2}\right) = -x$.

So the expression is

5. Absorb the constant …

Use $\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$.

Then the integrand is

$(1-3\sin^2x\cos^2x)+3\sin^2x\cos^2x=1.$ …

Step 1 — Set up with the binomial theorem.

Let $a = x^2$ and $b = y^2$. Then

Step 2 — Subtract term by term.

Now examine the bracket $\left[1-(-1)^r\right]$:

• If $r$ is even: $(-1)^r = +1$, so the bracket is $1 - 1 = 0$ — the term vanishes.
• If $r$ is odd: $(-1)^r = -1$, so the bracket is $1 - (-1) = 2$ — the term survives, doubled.

Step 3 — Count the surviving terms.

We need the odd $r$ with $0 \le r \le 25$:

Count: these form an AP with first term $1$, common difference $2$, last term $25$, so the number of terms is

$\frac{25-1}{2} + 1 = 12 + 1 = \mathbf{13}.$ …

Each entry is linear in $x$, so the determinant is a polynomial in $x$; its constant term is the value at $x=0$.

Setting $x=0$:

Expanding along the first row: …

The key insight here is that the integrand $\frac{1}{1+e^x}$ doesn't fit a basic formula directly, but a clever algebraic manipulation turns it into something we can integrate immediately. The trick is to multiply numerator and denominator by $e^{-x}$ — this transforms the expression into a form where the numerator is exactly the derivative of the denominator (up to a sign), which is the hallmark of a natural logarithm integral.

Let’s work through it step by step.

1. Rewrite the integrand Multiply numerator and denominator by $e^{-x}$:

Notice that $e^{-x} + 1 = 1 + e^{-x}$, so we have:

2. Recognize the derivative pattern The derivative of $1 + e^{-x}$ is $-e^{-x}$. Our numerator is $e^{-x}$, which is exactly $-1$ times that derivative. So:

More directly, we can set $u = 1 + e^{-x}$, then $du = -e^{-x} dx$, so $e^{-x} dx = -du$. The integral becomes:

3. Rewrite the answer in terms of $e^x$ Since $e^{-x} = \frac{1}{e^x}$, we have:

Using the logarithm property $-\ln\left(\frac{A}{B}\right) = \ln\left(\frac{B}{A}\right)$, we get: