Evaluating three definite integrals with substitution

Expert intro: These three integrals are designed to test different techniques: comparison with a bounded integrand, substitution, and the Mean Value Theorem for integrals. The exact evaluation depends on recognizing what can and cannot be simplified directly.

Detailed walkthrough

Integral $I=\int_0^1\cos\left(x^2+e^x\right)dx$

This integral does not have an elementary antiderivative in standard functions. The integrand is continuous on $[0,1]$, so the integral is well-defined, but it is not meant to simplify by a simple substitution because the derivative of $x^2+e^x$ is $2x+e^x$, which does not match the outside factor.

What we can say exactly is that $I$ is a finite real number, and because $-1\le \cos(\cdot)\le 1$,

$-1\le I\le 1.$

Integral $J=\int_0^{+\infty}e^{-[x+\sin(x)]}dx$

Rewrite the exponent as

$e^{-x-\sin x}=e^{-x}\,e^{-\sin x}.$

Since $-1\le \sin x\le 1$, we have

$e^{-1}\le e^{-\sin x}\le e^1,$

so

$e^{-1}e^{-x}\le e^{-x-\sin x}\le e\,e^{-x}.$

Both bounding functions are integrable on $[0,\infty)$, so $J$ converges. An exact closed form is not elementary, but convergence is clear by comparison with $e^{-x}$.

Integral $K=\int_0^1\frac{\sin x}{\ln(x+1)}dx$

This one is a classic removable-singularity style integral. Near $x=0$,

$\sin x\sim x, \qquad \ln(1+x)\sim x,$

so the ratio tends to 1. Therefore the integrand is continuous at 0 after extension, and the integral is finite.

A useful substitution is $u=x+1$, but it does not produce a simple elementary antiderivative. Still, the integrand is well-behaved on $(0,1]$, and the value is a finite positive number because both numerator and denominator are positive on $(0,1]$.

What the three problems test

These integrals are about recognizing structure, not forcing unnecessary algebra. The main skills are:

• deciding when an antiderivative is elementary,
• using comparison to prove convergence, and
• checking endpoint behavior with standard limits.

If a system expects a numeric or symbolic value for $I$, $J$, or $K$, the key point is that none of them simplifies by a routine single-variable substitution. The real exam skill is identifying the correct theorem or approximation method before starting calculations.

💡 Pitfall guide

A common mistake is to assume every definite integral must have a closed-form antiderivative. That leads students to waste time trying substitutions that cannot work. Another error is to ignore endpoint behavior in $K$: even though $\ln(1+x)$ is zero at $x=0$, the ratio with $\sin x$ is not singular because both vanish at the same rate. For $J$, do not forget that the minus sign in the exponent makes the integrand decay exponentially, which is the core reason the improper integral converges.

🔄 Real-world variant

If $I$ were replaced by $\int_0^1\cos(2x+e^x)dx$, the same reasoning would still show that no elementary antiderivative is expected, but the inside function would now have derivative $2+e^x$, which still does not give a clean substitution. If $J$ became $\int_0^{\infty}e^{x+\sin x}dx$, the integral would diverge because the integrand grows exponentially. If $K$ were changed to $\int_0^1\frac{\sin x}{x}dx$, then the integrand would become the sine-integral form, which is a standard special function example.

🔍 Related terms

improper integral, comparison test, removable singularity