1. Given

$f(x)=7\tan^8 x+7\tan^6 x-3\tan^4 x-3\tan^2 x$

and

$I_1=\int_0^{\pi/4} f(x)\,dx,\qquad I_2=\int_0^{\pi/4} x f(x)\,dx.$

We need to find:

$7I_1+12I_2.$

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2. Factorize $f(x)$

Let $t=\tan^2 x$. Then

$f(x)=7t^4+7t^3-3t^2-3t.$

Factor:

$f(x)=t(7t^3+7t^2-3t-3).$

Now,

$7t^3+7t^2-3t-3=(t+1)(7t^2-3).$

So,

$f(x)=t(t+1)(7t^2-3).$

Substituting back $t=\tan^2 x$ and using $1+\tan^2 x=\sec^2 x$,

$f(x)=\tan^2 x\sec^2 x(7\tan^4 x-3).$

This suggests the substitution $u=\tan x$.

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3. Compute $I_1$

Using $u=\tan x$, so that

$du=\sec^2 x\,dx,$

and when $x=0$, $u=0$; when $x=\pi/4$, $u=1$.

Thus,

=\int_0^1 u^2(7u^4-3)\,du.$$ So, $$I_1=\int_0^1 (7u^6-3u^2)\,du =\left[u^7-u^3\right]_0^1=1-1=0.$$ Hence, $$I_1=0.$$ --- 4. **Use symmetry to compute $I_2$** We use the transformation $x\mapsto \frac{\pi}{4}-x$. First compute $f\left(\frac{\pi}{4}-x\right)$. Using $$\tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x},$$ this direct substitution is messy. Instead, let us rewrite $f(x)$ in terms of $t=\tan x$ and test a useful relation. We already have $$f(x)=7t^8+7t^6-3t^4-3t^2.$$ Now let $$s=\tan\left(\frac{\pi}{4}-x\right)=\frac{1-t}{1+t}.$$ A simpler route is to guess and verify the symmetry relation needed for weighted integrals: $$f\left(\frac{\pi}{4}-x\right)=f(x).$$ But checking numerically at $x=0$ gives $$f(0)=0,$$ while at $x=\frac{\pi}{4}$, $$f\left(\frac{\pi}{4}\right)=7+7-3-3=8,$$ so this is false. Instead, let us compute $I_2$ directly by integration by parts. --- 5. **Compute $I_2$ by parts** Since $$f(x)=\tan^2 x\sec^2 x(7\tan^4 x-3),$$ observe that if $$F(x)=\tan^7 x-\tan^3 x,$$ then $$F'(x)=7\tan^6 x\sec^2 x-3\tan^2 x\sec^2 x =\tan^2 x\sec^2 x(7\tan^4 x-3)=f(x).$$ So, $$f(x)=\frac{d}{dx}(\tan^7 x-\tan^3 x).$$ Thus, $$I_2=\int_0^{\pi/4} x f(x)\,dx =\int_0^{\pi/4} x F'(x)\,dx.$$ Integrating by parts: $$\int xF'(x)\,dx=xF(x)-\int F(x)\,dx.$$ Therefore, $$I_2=\left[x(\tan^7 x-\tan^3 x)\right]_0^{\pi/4}-\int_0^{\pi/4}(\tan^7 x-\tan^3 x)\,dx.$$ At $x=\pi/4$, $\tan x=1$, so $$\tan^7 x-\tan^3 x=1-1=0.$$ At $x=0$, it is also $0$. Hence boundary term is zero, and $$I_2=-\int_0^{\pi/4}(\tan^7 x-\tan^3 x)\,dx.

So,

$I_2=\int_0^{\pi/4}(\tan^3 x-\tan^7 x)\,dx.$

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6. Evaluate the integrals

Use $t=\tan x$, $dx=\dfrac{dt}{1+t^2}$, with limits $0$ to $1$.

Then

$I_2=\int_0^1 \frac{t^3-t^7}{1+t^2}\,dt.$

Simplify:

$\frac{t^3-t^7}{1+t^2}=\frac{t^3(1-t^4)}{1+t^2}=t^3(1-t^2)=t^3-t^5.$

Therefore,

=\left[\frac{t^4}{4}-\frac{t^6}{6}\right]_0^1 =\frac14-\frac16=\frac1{12}.$$ So, $$I_2=\frac1{12}.$$ --- 7. **Compute $7I_1+12I_2$** Since $I_1=0$ and $I_2=\dfrac1{12}$, $$7I_1+12I_2=7(0)+12\left(\frac1{12}\right)=1.$$ --- 8. **Option check** - A: $2\pi$ ❌ - B: $1$ ✅ - C: $\pi$ ❌ - D: $2$ ❌ Therefore, the correct option is **B**.